Contrapositively, suppose $\{V_n\}$ is a cover with no finite subcover. Then for $W_n = \bigcup_{m \le n} V_n$, each $X \setminus W_n$ is nonempty. So for each $n$, choose $x_n \in X \setminus W_n$. The sequence $\{x_n\}$ cannot have a converging subsequence: If $p \in X$, then $p \in V_m$ for some $m$ and so $x_n \notin V_m$ for all $n > m$.

Let $\{K_n\}$ be a countable compact cover of $X$ of which for all $x \in X$, some $K_n$ is a compact nbd of $x$. Let $K$ be compact. For each $x \in K$, let $n_x$ be the least element of which $K_n$ is a compact nbd. Then $\{\text{int}(K_{n_x})\}$ is a cover of $K$, which must have a finite subcover, and so finitely many $K_n$. Meaning $A \subseteq \bigcup_{n \le m} K_n$ for big enough $m$.

So define $K'_m = \bigcup_{n \le m} K_n$. Clearly $\{K'_m\}$ is a countable compact cover of $X$ and any compact is contained in some $K'_m$ by above.

We just have to argue a $k$-netork is a network: Let $\mathcal{N}$ be a $k$-network. Singletons are compact. So in a $k$-network, for each $x$ in an open set $U$, we can find $\mathcal{N}^*_x \subseteq \mathcal{N}$ with $\{x\} \subseteq \bigcup \mathcal{N}^*_x \subseteq U$ (we don’t need $\mathcal{N}^*_x$ to be finite here, so no need for the axiom of choice, just take every set of $\mathcal{N}$ containing $x$). Thus,

$$U = \bigcup_{x \in U} \{x\} \subseteq \bigcup_{x \in U} \mathcal{N}^*_x \subseteq U \ \implies \ \text{$U$ is a union of sets in $\mathcal{N}$, so it is a network}$$

Let $A$ be an infinite set. Contrapositively, suppose every limit point of $A$ is not an $\omega$-accumulation point. If $p$ were to be a limit point of $A$, there exists a nbd $V$ of $p$ of which $V \cap A = \{ a_1, \dots, a_n \}$ is finite. But now for each $a_k$, there’s $V_k$, a nbd of $p$, so that $a_k \notin V_k$ (by being $T_1$). Then $W = V \cap V_1 \cap \cdots \cap V_n$ is a nbd of $p$ with $W \cap A = \emptyset$. So $A$ is an infinite set with no limit points.

Let $\mathcal{V}$ be a collection of pairwise-disjoint nonempty open sets. Let $D$ be a dense countable set. For each $V \in \mathcal{V}$, $D \cap V \ne \emptyset$ means it must be the nbd of some $x_n \in D$. So let $n \sim V$ if and only if $x_n \in V$. As the collection is pairwise-disjoint, this induces a function $f : \omega \to \mathcal{V}$ with $f(n) = V$ that must be surjective, and so $\text{card}(\omega) \ge \text{card}(\mathcal{V})$.

Let $\mathcal{N} = \{ \{x\} \ : \ x \in X \}$ be the network of singletons. The open sets don’t matter: If any compact $K$ is necessarily finite, then $N^* = \{ \{x\} \ : \ x \in K \} \subseteq \mathcal{N}$ is finite and $K \subseteq N^*$. Since $X$ is countable, so is $\mathcal{N}$.

Let $x \ne y$. Since $X$ is $T_2$, let $x \in U$ and $y \in V$ nbds with $U \cap V = \emptyset$. This means $x \notin \overline{V}$ and $y \notin \overline{U}$, so being regular, let $W_x,W_V$ be disjoint nbds with $x \in W_x$ and $\overline{V} \subseteq W_V$ and similarly $W_y,W_U$ disjoint nbds with $y \in W_y$ and $\overline{U} \subseteq W_U$. Then $O_x = W_x \cap U$ and $O_y = W_y$ are the nbds of $x$ and $y$ of which $\overline{O_x} \cap \overline{O_y} = \emptyset$ (remember that $\overline{O_x} \subseteq \overline{W_x} \cap \overline{U}$).

Let $A$ be a closed set and $x \notin A$. We argue $\overline{\{x\}} \cap A = \emptyset$: Every point of $\overline{\{x\}}$ is indistinguishable from $x$, as the space is $R_0$, so $y \in \overline{\{x\}} \cap A$ would imply $x \in \overline{A} \subseteq A$. So by Urysohn’s lemma, let $f : X \to [0, 1]$ with $f(A) = \{0\}$ and $f(\overline{\{x\}}) = \{1\}$. In particular, $f(x) = 1$.

Let $C$ be a connected component, and $x$ an isolated point of $C$. $X$ is $T_1$ and $C$ is closed, so $\{x\}$ is closed in $C$, and take $W$ the nbd of $x$ with $W \cap C = \{x\}$. Then $C \setminus \{x\}$ and $W$ is a separation of $C$.

Every open set is connected: If $V$ were not to be, then $V = U_1 \cup U_2$ with $U_1,U_2$ disjoint and open in $V$, but then they must be open in $X$ as well.

By definition, it suffices to show the space is $\sigma$-compact. Let $\mathcal{K}$ be the set of all compact neighborhoods. So $\{\text{int}(K)\}_{K \in \mathcal{K}}$ forms an open cover. Let $\mathcal{C} \subseteq \mathcal{K}$ be countable such that $\{\text{int}(K)\}_{K \in \mathcal{C}}$ is a countable subcover. Then $X = \bigcup_{K \in \mathcal{C}} K$.

Let $A$ be closed and $x \notin A$. Then $V \cap A = \emptyset$ for some nbd of $x$. Let $B \subseteq V$ be a clopen nbd of $x$. Then $\chi_B$ (indicator function of $B$) is continuous, as any preimage is one of $\{\emptyset, B, B^c, X\}$, with $\chi_B(A) = \{0\}$ and $\chi_B(x) = 1$.

A LOTS space is clearly $T_1$. For $x < y$, consider the nbds $({\leftarrow}, y)$ and $(x, {\rightarrow})$. For any given $x$, if $x = \min X$, the sets $\{ [x, y) \ : \ y > x \}$ are open. If $x = \max X$, the sets $\{ (y, x] \ : \ y < x \}$ are open. Otherwise, $\{ (y, z) \ : \ y < x < z \}$ are open. In all three cases, we have a local basis for $x$ which is trivially order-convex (unless $X = \{x\}$, but then the result is trivial).

For each $x$, let $V$ be a nbd with $f : V \to \R^n$ a homeomorphism to $f(V)$. Let $\delta$ be small enough such that $B = B(f(x), \delta) \subseteq f(V)$. Any open ball is arc connected (consider $h(t) = tx + (1-t)y$), so $B_n = B(f(x), \delta/n)$ is a local basis of arc connected sets within $B$ which means $U_n = f^{-1}(B_n)$ have the same property (homeomorphisms preserve local bases and arc-connectedness). Now do this with every $x$ to form a basis.

Let $f : X \to [0, 1]$ with $f(a) = 0$ and $f(b) = 1$. Take $V_a = f^{-1}([0, 1/3))$ and $V_b = f^{-1}((2/3, 1])$. Then $f(\overline{V_a}) \subseteq \overline{f(V_a)} \subseteq [0, 1/3]$ and $f(\overline{V_b}) \subseteq \overline{f(V_b)} = [2/3, 1]$. This proves $\overline{V_a} \cap \overline{V_b} = \emptyset$.

Let $\{N_n\}$ be the countable network and $A \subseteq X$. Ignore the network sets that don’t intersect $A$ and choose $x_n \in N_n \cap A$. Then $\{x_n\}$ has to be dense in $A$: For any $x \in A \cap V$, $x \in N_n \cap A \subseteq A \cap V$ for some $n$ so that $x_n \in A \cap V$.

If any path connected component is open, then there can only be one (as they form a partition of $X$, disproving being connected). Take the path connected component $P$ of $x$. For $y \in P$, let $U$ be a connected nbd. Then any $z \in U$ is path connected to $x$ (just concatenate the paths $x \sim y$ and $y \sim z$). So $U \subseteq P$.

Let $f : X \to Y \subseteq \R$ be a homeomorphism. $\R$ is $T_1$, so $Y \simeq X$ is as well. Define order on $X$ by $x < y$ iff $f(x) < f(y)$. If $U$ is a nbd of $x$, then $f(x) \in f(U)$ and $f(U) = V \cap Y$, where $I = (f(x) - \delta, f(x) + \delta) \subseteq V$ for some $\delta > 0$. Let $J = f^{-1}(I) \subseteq U$. Suppose $a,b \in J$ and $a < z < b$. Then $f(a) < f(z) < f(b)$ with $f(a),f(b) \in I$, so $f(z) \in I$ and $z \in J$ as we wished, to prove $J$ is order-convex.

A GO-space has a basis of order-convex neighborhoods. A LOTS space has a basis containing all open intervals. So by contraposition, suppose $A$ is some order-convex neighborhood of a point $x$ that contains no open intervals. Then it must be closed. If $y \in \cl{A}$ and $y < x$, then there is some $z \in (y, x) \cap A$. But then either $(z, x) \subseteq A$ or $(x, z) \subseteq A$. The same holds if $y > x$. But then $A$ is a proper nonempty clopen set, and $X$ is not connected (unless $X = \{x\}$, but then the result is trivial).

It suffices to show a basis is a $k$-network: If $K \subseteq U$ with $U$ open and $K$ compact, then $U = \bigcup \mathcal{B}$, where $\mathcal{B}$ is a set of basis elements. $K$ being compact means there exists a finite subset $\mathcal{B}^* \subseteq \mathcal{B}$ with $K \subseteq \bigcup \mathcal{B}^* \subseteq U$.

It suffices to show that for $x \in W$ with $W$ open, there is some regular open set $O$ with $x \in O \subseteq W$. By regularity, let $x \in U$ and $W^C \subseteq V$ open sets with $U \cap V = \emptyset$. In particular, $\cl{U} \cap W^C = \emptyset$, so $\cl{U} \subseteq W$. Then just pick $O = \text{int}(\cl{U})$.

Let $f : V \to \R^n$ be an embedding with $V$ nbd of $x$. If $n = 0$ for every choice of $x$, the space is discrete with two or more points, so clearly not hyperconnected. If $n \ge 1$ for some $x$, then $f(V)$ is an open set with infinitely many points, so by taking any two points with disjoint nbds $U$ and $V$ ($\R^n$ is $T_2$), we have $f^{-1}(U)$ and $f^{-1}(V)$ disjoint, open, nonempty.

Let $U$ be a cozero set. Then by perfect normality, let $f : X \to [0, 1)$ be continuous with $f^{-1}(\{0\}) = \cl{U}$ (since $\emptyset$ is closed, so we chose it as the set where $f^{-1}(\{1\}) = \emptyset$, this is in fact why every closed set is a zero set in perfectly normal spaces). It follows that $V = f^{-1}((0, 1))$ is cozero with $U \cap V = \emptyset$ and $X = \cl{U} \cup V \subseteq \cl{U \cup V}$.

Let $p \in \cl{A}$ and choose $B$ a countable basis nbd of $p$. Then $D = A \cap B$ is a countable subset of $A$ such that if $V$ is any open nbd of $p$, $B \cap V$ is open, so $A \cap (B \cap V) = D \cap V \ne \emptyset$.

Take any $x$ and a nbd $U$ which is $T_2$. We wish to find a neighborhood $V$ not containing some other $y$. If $y \notin U$, just take $V = U$. Otherwise, take $x \in V \subseteq U$ with $y \notin V$ since $U$ is $T_1$. But then $V$ is also open in $X$, so we’re done.

If $A$ and $B$ are disjoint closed sets, then $\cl{A} \cap \cl{B} = \emptyset$ and so $\{A, B\}$ is a discrete family.

Take $x \in W \subseteq K$ with $V$ open and $K$ compact. Let $U$ and $V$ be a separation of $x$ and the closed set $W^C$. In particular, $\cl{U} \subseteq K$. The sets of the form $U \cap O$ where $O$ is open forms a local basis for $x$ and each closure $\cl{U \cap O} \subseteq K$ is compact.

Fix $p \in X$. If $p$ is isolated, $\{\{p\}\}$ is a local basis. Otherwise, $p \in \cl{X \setminus \{p\}}$ and there is a countable $D \subseteq X \setminus \{p\}$ with $p \in \cl{D}$. So consider the intervals of the form $[p, x)$, $(x, p]$, or $(x, y)$ for $x,y \in D$ which are open in $X$. This must be countable and it is a local basis of $p$ since $p \in \cl{D}$ and it isn’t isolated in $X$.

The proof is literally the same as (T224). I’d like to express my discontent with $\pi$-base’s usage of the term “local basis”. To me, a local basis is by definition a collection of open sets, but what they mean by “locally compact” is: Each $x$ has a collection of compact sets whose interiors form a local basis for $x$.

It suffices to prove any $F_\sigma$ set is Lindelöf, as that implies every open set is Lindelöf in a $G_\delta$ space. This follows from two basic topological facts: (1) Closed sets of Lindelöf spaces are Lindelöf: If $A \subseteq X$ is closed and $\mathcal{U}$ is a cover of $A$, then $\mathcal{U} \cup \{A^C\}$ is a cover of $X$ with a countable subcover, which induces a countable subcover of $A$ (2) A countable union of Lindelöf spaces $\{L_n\}$ is Lindelöf: If $\mathcal{U}$ is a cover of $L = \bigcup_{n < \omega} L_n$, then it is a cover of each $L_n$, which gives us a countable subcover $\mathcal{U}_n$ of $L_n$, and so $\bigcup_{n < \omega} \mathcal{U}_n$ is a countable subcover of $L$.

Let $A$ be closed. Then there is a continuous map $f : X \to [0, 1]$ with $f^{-1}(\{0\}) = A$. Thus, each $V_n = f^{-1}([0, 1/n))$ is open and $A = \bigcap_{n=1}^\infty V_n$ is a $G_\delta$ set.

Let $\{N_k\}$ be the countable network. Let $A \subseteq X$ and $\mathcal{U}$ be an open cover of $A$. We know for any $x \in U \in \mathcal{U}$ there is a $k$ such that $x \in N_k \subseteq U$. Let $I$ be the set of $k < \omega$ such that $N_k \subseteq U$ for some $U \in \mathcal{U}$, and for each $k \in I$, choose one such $U_k$. Then $\{U_k\}_{k \in I}$ is a countable subcover.

In $R_0$ spaces, points $x,y$ are indistinguishable iff $\cl{\{x\}} = \cl{\{y\}}$. Thus, if $x \in V$ and $V$ is open, $\cl{\{x\}} \subseteq V$. So $V = \bigcup_{x \in X} \cl{\{x\}}$ is an $F_\sigma$ set, as $X$ is countable.

Let $\mathcal{U}$ be a $k$-cover and $\mathcal{B}$ a countable basis. For any compact $K$, $K \subseteq U$ for some $U \in \mathcal{U}$, which can be written as a union of basis elements $U = \bigcup \mathcal{B}^*$. This is an open cover of $K$, so we can extract a finite subcover. This allows to conclude that if for each finite selection of basis elements, we select one element $U$ which contains their union, then this must be a $k$-subcover, and the collection of all finite selections of basis elements $\bigcup_{n < \omega} \mathcal{B}^n$ is countable.

$x$ and $y$ are indistinguishable iff either $x \notin \cl{\{y\}}$ or $y \notin \cl{\{x\}}$. Suppose the former. Then regularity gives us $x \in U$ and $\cl{\{y\}} \subseteq V$ with $U \cap V = \emptyset$, which is clearly a separation of $x$ and $y$.

To prove $R_0$, it suffices to prove $x \notin \cl{\{y\}} \implies y \notin \cl{\{x\}}$. Note that $\cl{\{y\}} = \bigcap_{n < \omega} V_n$ is a $G_\delta$ set, so there exists some $n$ such that $x \notin V_n$, yet $y \in V_n$. That proves $y \notin \cl{\{x\}}$.

A countable union of subsets in a finite space can be assumed to be a finite union by removing duplicates. Now we just use the fact that for finitely many closed sets $\{A_1, \dots, A_n\}$, that $\text{int}(\bigcap_{k=1}^n A_i) = \bigcap_{k=1}^n \text{int}(A_i)$. That is, a finite union of nowhere dense sets is nowhere dense.

Let $\mathcal{N}$ be a countable network. For each $x \in X$, define $f(x) = \{ N \in \mathcal{N} \ : \ x \in N \}$. If $x \ne y$, by $T_0$ and without loss of generality, assume $x \in U$ and $y \notin U$ for some open $U$. Since $U$ is a union of network elements, there is some $N$ with $x \in N$ yet $y \notin N$. Then $f : X \to \mathcal{P}(\mathcal{N})$ is injective, and so $\card{X} \le \card{\mathcal{P}(\mathcal{N})} = 2^{\aleph_0} = \mathfrak{c}$.

Let $\mathcal{B}$ be a basis of clopen sets and let $\mathcal{U}$ be an open cover. For each $U \in \mathcal{U}$, define $\mathcal{B}_U = \{ B \in \mathcal{B} \ : \ B \subseteq U \}$. Then $\mathcal{V} = \bigcup_{U \in \mathcal{U}} \mathcal{B}_U$ is an open refinement of clopen sets. By Lindelöf, let $\{B_n\}$ be a countable subrefinement of $\mathcal{V}$. If $B^*_n = B_n \setminus \bigcup_{k < n} B_k$, then $\{B^*_n\}$ are all clopen and pairwise disjoint, so it is a clopen refinement which partitions $X$.

Let $\mathcal{U}$ be an open cover. Let $\mathcal{F}$ be the collection of finite subsets of $\mathcal{U}$ and define a function $v(F) = \bigcup F$ on $\mathcal{F}$. Then $\{v(F)\}_{F \in \mathcal{F}}$ is an $\omega$-cover, since $\mathcal{U}$ is a cover of any finite (compact) set $K$, which has a finite subcover, meaning $K \subseteq \bigcup F$ for some finite $F \subseteq \mathcal{U}$. Let $\mathcal{A} \subseteq \mathcal{F}$ be countable so that $\{v(F)\}_{F \in \mathcal{A}}$ is a countable $\omega$-subcover. Then $\bigcup \mathcal{A} \subseteq \mathcal{U}$ is a countable subcover.

A locally-finite collection in a Lindelöf space is countable, seen in the proof of (T387), and so the network is a countable union of countable networks, hence, countable.

A locally-finite collection in a Lindelöf space is countable: If $\mathcal{A}$ is locally finite, the sets $\mathcal{U}$ which intersect finitely many elements of $\mathcal{A}$ is an open cover, so if $\mathcal{U}^*$ is a countable subcover, then every element of $\mathcal{A}$ intersects some $U \in \mathcal{U}^*$, yet only finitely many intersect each $U$, so $\mathcal{A}$ is countable.

Start by showing regularity. Let $x \notin A$ and $A$ closed. Construct the pairs $(U, V)$ of nbds such that $x \in U$ and $y \in V$ for some $y \in A$ with $U \cap V = \emptyset$. Then the sets $V$ form an open cover of $A$. Closed sets are Lindelöf, so take $\{(U_n, V_n)\}$ a countable subcollection with $\{V_n\}$ still covering $A$. Then $x \in U \coloneqq \bigcap_{n < \omega} U_n$ is open in a P-space, and $A \subseteq V \coloneqq \bigcup_{n < \omega} V_n$, with $U \cap V = \emptyset$.

Now let $A,B$ be closed and disjoint. Construct the pairs $(U, V)$ of nds such that $A \subseteq U$ and $x \in V$ for some $x \in B$ with $U \cap V = \emptyset$. Similarly, we take $\{(U_n, V_n)\}$ such that $B \subseteq V \coloneqq \bigcup_{n < \omega} V_n$ and $A \subseteq U \coloneqq \bigcap_{n < \omega} U_n$, which is open in a P-space, and $U \cap V = \emptyset$.

Given $x \ne y$, let $f : X \to [0, 1]$ such that $f(x) = 0$ and $f(y) = 1$. Then $V_n = f^{-1}([0, 1/n))$ are all open. Then $x \in B = \bigcap_{n=1}^\infty V_n$ is open in a P-space, and $B$ is closed, since $B = f^{-1}\left( \bigcap_{n=1}^\infty [0, 1/n) \right) = f^{-1}(0)$.

Contrapositively, let $f : X \to \R$ is continuous and non-constant. So $f(x) = \alpha$ and $f(y) = \beta$ with $\alpha \ne \beta$. Clearly $A_1 = f^{-1}(\alpha)$ and $A_2 = f^{-1}(\beta)$ are disjoint. They are zero sets, since $A_1 = g^{-1}(0)$ for $g(x) = f(x) - \alpha$ and similarly, $A_2$ is a zero set. So $A_1 \subseteq B_1$ and $A_2 \subseteq B_2$ with $B_1,B_2$ disjoint clopen sets, by strongly zero-dimensional. In particular, $x \in B_1 \subset X$, so the space is not connected.

If you remove “countable” from the definition of pseudonormal, that’s precisely one property of normal spaces. If $H$ is closed and $H \subseteq U$ is open, then $H \cap U^C = \emptyset$, so there are open sets $H \subseteq V$ and $U^C \subseteq V'$ with $V \cap V' = \emptyset$. In particular, $\cl{V} \subseteq U$.

Let $x \notin A$ with $A$ closed. Being $T_1$, $\{x\}$ is a countable closed set, and $\{x\} \subseteq U = A^C$ is open. So let $V$ be open with $x \in V \subseteq \cl{V} \subseteq A^C$. Then $A \subseteq \cl{V}^C = W$ with $V \cap W = \emptyset$.

Let $H = \{x_n\}$ and $H \subseteq U$ open. Then $x_n \notin F = U^C$ for each $n$, so by regularity, let $x \in V_n$ and $F \subseteq U_n$ with $V_n \cap U_n = \emptyset$. Then $A \subseteq V = \bigcup_{n < \omega} V_n$ and $F \subseteq W = \bigcap_{n < \omega} U_n$ with $V \cap W = \emptyset$. But that implies $\cl{V} \cap F = \emptyset$, so $H \subseteq V \subseteq \cl{V} \subseteq U$.

Let $f : X \to \R$ be injective. Using the usual topology of $\R$, we can define $d(x, y) = \abs{f(x) - f(y)}$ which is clearly a metric, and under this new metrizable topology for $X$, $f$ is a homeomorphism from $X$ to $f(X) \subseteq \R$, and any subset of $\R$ has a countable dense subset. This separable metrizable topology is trivially coarser than the discrete topology.

Let $(x_n)$ be a sequence with $x_n \to x$. Then $A = \{ x_n \ : \ n < \omega\} \cup \{x\}$ is compact, since for any open cover of $A$, some element $V$ nbd of $x$ must contain $x_n$ for all $n \ge n_0$ and $\{ x_n \ : \ n < n_0 \}$ is finite. But then $A$ is finite, and in a $T_1$ space, it must be discrete. So for some $n_0$, $n \ge n_0 \implies x_n \in \{x\} \implies x_n = x$.

All countable sets are closed, since they are a countable union of singletons (closed by $T_1$). If $A$ is countable and some $p \in A$ were not to be isolated, then $p \in \cl{A \setminus \{p\}}$, which would imply $A \setminus \{p\}$ is countable yet not closed.

Contrapositively, if $p$ is not an isolated point, then $p \in \cl{X \setminus \{p\}}$ and then there exists a sequence $(x_n)$ in $X \setminus \{p\}$ with $x_n \to p$. But then this sequence cannot be eventually constant, since the constant would have to be $p$ itself.

Let $A$ countable with $\cl{A} = X$. For any $x \in X$, $A \cup \{x\}$ is countable, so it is discrete. Since $x \in \cl{A}$, we have $x \in A$ and so $X = A$ is countable.

Contrapositively, if $p$ is not an isolated point, then $p \in \cl{X \setminus \{p\}}$, so there exists a countable $D \subseteq X \setminus \{p\}$ with $p \in \cl{D}$. But then $D \cup \{p\}$ is countable yet not discrete.

Let $x \in U = \text{int}(\cl{U})$ with $y \notin U$. If $U$ were to be dense, we’d have $\cl{U} = X$, but $\text{int}(\cl{U}) = U \ne X = \text{int}(X)$.

Let $x \in U \subseteq K$ and $y \in V \subseteq K'$ with $U,V$ open, $K,K'$ compact, and $x \ne y$. Then $L = K \cup K'$ is compact, so it is $T_2$. Thus, let $x \in W$ and $y \in W'$ be nbds such that $W \cap W' \cap L = \emptyset$. Then $(U \cap W) \cap (V \cap W') = \emptyset$ separates $x$ and $y$ by open sets in $X$.

Let $K \subseteq X$ compact and $x \notin K$. Then $K \cup \{x\}$ is compact and $T_2$. $T_2$ spaces separate points from compact sets, so there exists a nbd $V$ of $x$ with $V \cap K = \emptyset$. So $K$ is closed.

If $X$ is disconnected, let $A$ be a clopen set with $p \in A$ and $q \notin A$. Then the function $f : X \to X$ defined as $f(x) = q$ at $x \in A$ and $f(x) = p$ at $x \notin A$, then $f$ is continuous yet has no fixed point.

Denote $x \sim y$ to mean $x$ and $y$ are indistinguishable. For each $x$, let $W(x)$ be its smallest nbd. Clearly $x \sim y \implies y \in W(x)$ by definition, and for each $y \not\sim x$, there is a nbd $V$ of $x$ with $y \notin V$, so $y \notin W(x)$. Furthermore, this proves that $\sim$ is an equivalence relation, since $x \sim y \iff W(x) = W(y)$ and the sets $[x]_{\sim} = W(x)$ form a partition of $X$.

A hyperconnected set cannot have two isolated points. Let $p$ be the only non-isolated point. Then the hyperconnected sets are only the ones of the form $A = \{x\}$ and $B = \{p, x\}$ with $p \in \cl{\{x\}}$ for some $x$. Suppose they’re closed. Then clearly $A = \cl{\{x\}}$, and $B$ is a closed subset of $\cl{\{x\}}$ that contains $x$, so $B = \cl{\{x\}}$.

Let $p$ be non-isolated with $X \setminus \{p\}$ discrete. Let $A$ be countably compact. If $A \subseteq X \setminus \{p\}$, then $A$ is finite, so it is closed by $T_1$. If $p \in A$, then for any $x \notin A$, $\{x\} \subseteq A^C$ is a nbd, so $A$ is closed.

If $C$ is hyperconnected, it’s in particular connected, so it’s a singleton. Evidently it must be the closure of itself (connected components are closed).

Let $x \sim y$ iff they are indistinguishable. The equivalent classes $[x]$ form a basis for a topology which must be finer than $X$ (if $U$ is a nbd of $x$, it must contain all $y \sim x$). It suffices to show each $[x]$ is an open set. But $Y = X \setminus [x]$ is compact, and any points of $Y$ are distinguishable from $x$, so it’s possible to find $x \in U$ and $Y \subseteq V$ nbds with $U \cap V = \emptyset$ (this is analogous to the result that for $T_2$ spaces, any point $x$ and a compact $K$ can be separated)