Separable means Density$= \aleph_0$.

$\cl{X} = X$

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).

If it has a closed point… it has… a point.

A space is a subspace of itself.

Any set in any local basis is compact.

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)

Take a basis of open injectively path connected sets. There’s a point $p$ of which the only possible basis element containing $p$ is $X$. So $X$ is injectively path connected.

Let $\mathcal{N} = \{ \{x\} \ : \ x \in X \}$ be the network of singletons. $\mathcal{N}$ is locally finite: Every point has a finite nbd, so it obviously intersects finitely many elements of $\mathcal{N}$.

Let $A \subseteq X$ and $p \in X$ such that $X \setminus \{p\}$ is discrete. Then all points of $A$ are isolated except for potentially $p$. That is, $\overline{A} \setminus A \subseteq \{p\}$. We wish to show $\overline{A} \subseteq \text{scl}(A)$ where scl denotes the sequential closure. Clearly $A \subseteq \text{scl}(A)$. If $A$ is not closed, then there is some sequence $(x_n)$ in $A$ such that $x_n \to x$ for some $x \in \overline{A} \setminus A$. But $x$ must be necessarily $p$, so $p \in \text{scl}(A)$.

Every countably compact set is also Lindelöf, so it is compact (T106), hence it is closed.