Nice. Let $A$ be a set such that for any $K \subseteq X$ compact and $T_2$, $A \cap K$ is open in $K$. It suffices to show $A$ is open in $X$. We wish to prove so using the $k_2$-space property, so let $K$ be any compact $T_2$ space and $f : K \to X$ any continuous map. $f(K)$ is immediately compact, so our objective is to prove it is $T_2$. Once that’s done, that means $A \cap f(K)$ is open, and so $f^{-1}(A) = f^{-1}(A \cap f(K))$ is open, proving $A$ is open.

$X$ is Weak Hausdorff, so $f(K)$ is closed. Furthermore, for any $x \in X$, the inclusion map $\text{id} : \{x\} \to X$ is continuous and $\{x\}$ is trivially compact and $T_2$, so $f(\{x\}) = \{x\}$ is closed, proving $X$ is $T_1$, and consequently, $f(K)$ is as well. We now verify that $f : K \to f(K)$ is a closed map: If $A \subseteq K$ is closed in $K$, since it must be also compact and $T_2$, $f(A)$ is closed in $X$, hence, in $f(K)$ as well.

Now take $a \ne b$ in $f(K)$. Being $T_1$, we know $A = f^{-1}(\{a\})$ and $B = f^{-1}(\{b\})$ are closed sets. They’re compact, and $K$ is $T_2$, so let $A \subseteq U$ and $B \subseteq V$ with $U \cap V = \emptyset$. Then let $U' = f(K) \setminus f(U^C)$ and $V' = f(K) \setminus f(V^C)$. Being a closed map, these are open sets of $f(K)$. Clearly $a \in U'$ and $b \in V'$. $z \in U' \cap V'$ would imply $z = f(w)$ with $w \in U \cap V$, so $U' \cap V' = \emptyset$.

Source: The category of CGWH spaces, N. P. Strickland

Let $\mathcal{R} = \{R_j\}$ be a point-countable open refinement of some open cover $\{U_i\}$. Contrapositively, suppose no countable subcover exists. Recursively, define $x_0$ as any point and $\mathcal{R}_0$ the countable collection of sets in $\mathcal{R}$ containing $x$. Then $X \setminus \mathcal{R}_0$ is nonempty and we can choose $x_1$ from that. Inductively, let $\mathcal{R}_n \subseteq \mathcal{R}$ of the sets containing $x_n$, so that $\bigcup_{j=0}^n \mathcal{R}_j$ is countable and so $x_{n+1} \in X \setminus (\bigcup_{j=0}^n \mathcal{R}_j)$.

The sequence $Y = \{x_n\}$ is infinite and has no cluster point: If $x$ were to be, take $x \in R_j$ and the minimum $n$ such that $x_n \in R_j$. By construction, it is the only element of $Y$ in $R_j$. For any sequence with no cluster point, we can construct a countable cover with no finite subcover. Take $U_0$ nbd of $x_0$, let $m$ be the least element of which $x_m \notin U_0$ and $U_1$ a nbd of $x_m$. Then if $m$ is the least element with $x_m \notin \bigcup_{k \le n} U_k$, let $U_{n+1}$ a nbd of $x_m$. Finally, if $V = X \setminus \bigcup U_n$, then $\{V\} \cup \{U_n\}$ is a countable cover of $X$ with no finite subcover.

By contradiction, suppose $x \ne y$ are topologically indistinguishable. Then $Y = X \setminus \{x, y\}$ must be disconnected, since $\{x, y\}$ is connected and $Y$ has at least two points. Then there are open sets $U,V$ such that $Y \subseteq U \cup V$ and $U \cap V \cap Y = \emptyset$. Note that $U$ can’t contain any of $x$ or $y$, otherwise it must contain both and $U$ and $V$ are a separation of $X$, which should be connected. Same goes for $V$. So $\{x, y\} = (U \cup V)^C$. But notice $U \cup \{x\}$ is connected: If $A,B$ is a separation by open sets with $x \in A$, then $y \in A$ and the pair $A \cup V$,$B \cap U$ is a separation of $X$. Through a similar reasoning, $V \cup \{y\}$ is connected. But then $X$ is not biconnected.

It suffices to prove Weak Hausdorff. Thus, let $K$ be compact $T_2$ and $f : K \to X$ continuous. Using the $k_2$-space property, we wish to show that for any other $K'$ compact $T_2$ and $g : K' \to X$ continuous, that $g^{-1}(f(K))$ is closed. Let $\Delta$ be the diagonal of $X^2$. If we define the function $f \times g : K \times K' \to X^2$ as $(f \times g)(x, y) = (f(x), g(x))$, then the set $L = (f \times g)^{-1}(\Delta)$ is closed, by the $k_2$-Hausdorff property ($K \times K'$ is compact $T_2$). If $\pi_2 : K \times K' \to K'$ is the projection map, since $K$ is compact, $\pi_2$ is a closed map. But $\pi_2(L) = g^{-1}(f(K))$.

Given $p \in \overline{A}$, choose $D \subseteq A$ countable with $p \in \overline{D}$. Suppose $(x_\alpha)_{\alpha < \lambda}$ is some transfinite sequence in $D$ with $x_\alpha \to p$. We now construct a sequence in $D$ which converges to $p$ and we’re done (thanks to PatrickR).

If some $y \in D$ is cofinal in $(x_\alpha)$, this means any neighborhood $U$ of $x$ must contain $y$, so we can construct a constant sequence $y_n = y$ with $y_n \to p$. If no $y \in D$ is cofinal, then each $I_y = \{ \alpha < \lambda \ : \ x_\alpha = y \}$ is finite and $\lambda = \bigcup_{y \in D} I_y$ is a countable union of finite sets, so it is a countable ordinal. We can assume $\lambda$ to be a regular cardinal, so it is already a sequence.

Let $V$ be an order convex nbd of $p \in \cl{A}$. If $A = ({\leftarrow}, p)$ and $B = (p, {\rightarrow})$, then we could assume either $A \cap V$ has no maximum or $B \cap V$ has no minimum. Otherwise, then we could take $x = \max A$ and $y = \min B$ then $(x, y) \cap V = \{p\}$ would be a nbd of $p$ and so $p \in A$ (it would then suffice to take the constant sequence $x_\alpha = p$). Without loss of generality, say $A \cap V$ is infinite.

Using the axiom of choice, specifically Zorn’s lemma, we can take the collection of all well-ordered subsets of $A \cap V$, and any maximal element $W$ must be cofinal in $A \cap V$. Since this set is well-ordered, there is an isomorphism $f : \xi \to W$ from some ordinal $\xi$ and $x_\alpha = f(\alpha)$ is a sequence with $x_\alpha \to p$ by cofinality.

If $X \simeq Y \subseteq \R$, for each $x \in X$, we shall assume $x$ is not minimum or maximum element of $L$, so that it has a local basis of sets of the form $(a, b) \cap X$ for $a,b \in L$ (otherwise the proof is similar but the intervals are either $[x, b)$ or $(a, x]$). Then there must be some $c \in (a, x)$ with $c \notin X$, otherwise $(a, x) \subseteq X$ would be a path connected subset¹. Similarly, there must be some $d \in (x, b) \setminus X$ so that $x \in (c, d) \cap X = [c, d] \cap X$ is a clopen nbd of $x$ in $X$. So the clopen nbds of $X$ form a basis.

(1): This is the one time in the proof where we use the fact that we’re in $\R$, because connected $\iff$ path connected in $\R$. Otherwise, we could construct a very similar proof for (Totally disconnected ∧ GO-space) $\implies$ Zero dimensional

Suppose $A$ is countably compact yet not closed. $X$ is sequential, so there is a sequence $(x_n)$ in $A$ and $p \in \cl{A} \setminus A$ with $x_n \to p$. Then $(x_n)$ has some accumulation point $y \in A$. There cannot be infinitely many $n$ for which $x_n = y$, otherwise we could extract a subsequence with $x_{n_k} \to y$, and by US, that would imply $y = p$. Therefore, by taking a subsequence if necessary, we can assume $y \ne x_n$ for all $n$. But then $y \notin B = \{x_n \ : \ n < \omega \} \cup \{p\}$ yet $y \in \cl{B}$. So $B$ isn’t closed, which would imply there is some sequence $(y_n)$ in $B$ with $y_n \to q$ and $q \in \cl{B} \setminus B$. Just as before, there can only be finitely many $n$ with $y_n = x_m$ or $y_n = p$ for any $m$. So we could extract a subsequence $(y_{n_k})$ such that $y_{n_k} = x_{m_k}$ is also a subsequence of $(x_n)$ and so $y_{n_k} \to p = q$, a contradiction.

We prove any closed set $A$ is a zero set. Let $A = \bigcap_{n=1}^\infty V_n$ be a $G_\delta$ set. Without loss of generality, assume $\{V_n\}$ is decreasing. Define $A_n = V_n^C$, and by normality, let $f_n : X \to [0, 1]$ a function where $f_n(A) = \{0\}$ and $f_n(A_n) = \{1\}$. We define $f : X \to [0, 1]$ as

$$f(x) = \sum_{n=1}^\infty \frac{f_n(x)}{2^n}$$

It already has the property $f^{-1}(\{0\}) = A$ as we wished, since $f(x) = 0$ iff $f_n(x)$ for all $n$ and $x \notin A$ implies $x \in A_n$ for some $n$ and $f_n(x) \ne 0$. To prove continuity, we use the Weierstrass M-test. Each $g_n(x) = f_n(x)/2^n \le 2^{-n}$ and $\sum 2^{-n} = 1$ converges, so $\sum g_n(x)$ converges absolutely and uniformly. As each finite sum $\sum_{n=1}^N g_n(x)$ is continuous, their uniform limit $f$ is continuous.

We shall show $X$ is normal and $G_\delta$ (T257). The former is a result of (Regular ∧ Lindelöf) $\implies$ Normal, and the proof of this fact can be seen in the second-half of (T26). There, we merely use the fact that the space is regular and closed sets are Lindelöf, which is clearly true in a Lindelöf space.

Now let $A$ be closed. By regularity, for each $x \notin A$ there is a pair $x \in U$ and $A \subseteq V$ of open sets with $U \cap V = \emptyset$. By joining all such pairs $(U, V)$ into a collection $\mathcal{W}$, we see that $\pi_1(\mathcal{W})$ is a cover of $A^C$. By hereditarily Lindelöf, there is a countable subcollection $\{(U_n, V_n)\}$ in $\mathcal{W}$ which, for any $x \notin A$, there is some $n$ for which $n \notin V_n$. Conversely, $x \in A$ implies $x \notin V_n$ for all $n$. So $A = \bigcap_{n < \omega} V_n$ is a $G_\delta$ set.

Let $A = f^{-1}(0)$ and $B = g^{-1}(0)$ be disjoint cozero sets, where $f,g : X \to \R$ are continuous. Then the function

$$h(x) = \frac{\abs{f(x)}}{\abs{f(x) + g(x)}}$$

is continuous, since the denominator would only be zero at $x$ if $x \in A \cap B$. $h$ maps $X$ to $[0, 1]$. Furthermore, $A = h^{-1}(0)$ and $B = h^{-1}(1)$. Note that $\{h^{-1}(t)\}_{0 \le t \le 1}$ is an uncountable family, yet $X$ is countable, so $h^{-1}(r) = \emptyset$ for some $r$. Then $h^{-1}([0, r))$ and $h^{-1}((r, 1])$ are clopen disjoint sets of $X$ which separate $A$ and $B$.

Let $\{K_n\}_{n=1}^\infty$ be a cover of compact nbds with $K_n \subseteq \text{int}(K_{n+1})$. Let $\mathcal{U}$ be any open cover. For each $n$, there is a finite $\mathcal{F}_n \subseteq \mathcal{U}$ which covers $K_n$. Now we use the fact that each $K_n$ is closed by KC, and so $U_n = \text{int}(K_n) \setminus K_{n-1}$ are all open (to define it at $n=1$, denote $K_0 = \emptyset$). Thus, each $U_n$ is covered by the finite sets $\mathcal{F}_n$, and if we define $\mathcal{R}_n = \{ U_n \cap F \ : \ F \in \mathcal{F}_n \}$, we get that $\mathcal{R} = \bigcup_{n=1}^\infty \mathcal{R}_n$ is an open refinement, where for each $x$, take the smallest $n$ for which $x \in \text{int}(K_n)$ so that $x \in U_n$ and so $x$ intersects at least one and at most all (finitely many) elements of $\mathcal{R}_n$.

Initially, for each $x \in X$, if a sequence $(y_n)$ does not converge to $x$, then there exists a nbd $V$ of $x$ such that for all $n_0 \in \omega$, there exists some $n > n_0$ such that $y_n \notin V$. This means there are infinitely many $n$ for which $y_n \notin V$, and so we can construct a subsequence $(y_{n_k})$ such that (1) it doesn’t have a converging subsequence either and (2) $y_{n_k} \notin V$ for all $k$.

Let $X = \{x_n\}$. Contrapositively, let $\mathbf{y} = (y_n)$ be a sequence with no converging subsequence. Let $V_0$ be a nbd of $x_0$ and construct a subsequence $\mathbf{y}_0$ of $\mathbf{y}$ such that every term in $\mathbf{y}_0$ lies outside $V_0$. Inductively, given $\mathbf{y}_n$, choose $V_{n+1}$ a nbd of $x_{n+1}$, and let $\mathbf{y}_{n+1}$ be a subsequence of $\mathbf{y}_n$ which lies entirely outside of $V_{n+1}$. Note that by construction, each $\mathbf{y}_n$ is a subsequence of $\mathbf{y}_k$ for any $k < n$, so all of its terms lie outside of $\bigcup_{k \le n} V_k$. This way, we’ve constructed an open cover $\{V_n\}$ of $X$ with no finite subcover, and so $X$ is not compact.