Condition for Open Extension Space to be Separable

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $T^*_{\bar p} = \left({S^*_p, \tau^*_{\bar p}}\right)$ be the open extension space of $T$.

Then $T^*_{\bar p}$ is a separable space iff $T$ is.

Proof
Let $T = \left({S, \tau}\right)$ be a separable space.

Then there exists a countable subset $H \subseteq S$ which is everywhere dense in $T$.

That is, $H^- = S$ where $H^-$ is the closure of $H$ in $S$.

So in $T_{\bar p}^*$, $S\subseteq H^-=S^-\subseteq S^*_p$.

Hence $H^- = S$ or $H^- = S \cup \{p\} = S^*_p$.

From Topological Closure is Closed we deduced that $H^-$ is closed in $T_{\bar p}^*$.

From the definition of open extension topology, the closed sets of $T_{\bar p}^*$ are $S^*_p\setminus U$ where $U\in\tau$.

Since $\{p\}\notin \tau$, $S$ is not closed.

Thus it must be that $H^- = S^*_p$.

Hence $H$ is a countable subset $H \subseteq S^*_p$ which is everywhere dense in $T^*_{\bar p}$.

Now suppose $T^*_{\bar p} = \left({S^*_p, \tau}\right)$ is separable and that $T = \left({S, \tau}\right)$ is not separable.

Let $H \subseteq S^*_p$ be a countable subset which is dense in $T^*_{\bar p}$.

If $H \subseteq S$, then $H^- = S$ in $T$, but this cannot be since $T$ is not separable.

Then $p \in H$.

Consider the subset $V = H \setminus \{p\}$.

From Topological Closure is Closed we have that $V^-$ is closed in $T_{\bar p}^*$.

From the definition of open extension space, all non-empty closed sets contain the point $p$.

So $V \cup \{p\} = H \subseteq V^-$.

From Set Closure is Smallest Closed Set, $S^*_p = H^- \subseteq V^- \implies V$ is dense in $T_{\bar p}^*$.

But since $V \subseteq S$, $V$ is dense in $S$ too.

From Subset of Countably Infinite Set is Countable, $V$ is countable as it is a subset of a countable set.

Thus $T$ is separable, which is a contradiction.

Finally, our assumption is false: $T$ is indeed separable.