Uncountable Excluded Point Space is not Separable

Theorem
Let $T = \left({S, \tau_{\bar p}}\right)$ be an uncountable excluded point space.

Then $T$ is not separable.

Proof
Let $H \subseteq S$ such that $H$ is countable.

Then $H \ne S$ as $S$ is uncountable by hypothesis.

From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$.

So, by definition, the closure of $H$ is $H \cup \left\{{p}\right\}$.

From Union of Countable Sets we have that $H \cup \left\{{p}\right\}$ is countable.

So $H \cup \left\{{p}\right\} \ne S$.

So $H$ is not everywhere dense in $T$.

Thus $T$ can have no countable subset of $S$ which is everywhere dense in $T$.

Hence the result by definition of separable space.