Countable Excluded Point Space is Separable

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

Then $T$ is a separable space.

Proof
The closure of the set $S$ (trivially) equals $S$.

That is, $S$ is everywhere dense in $T$.

But as $S$ is countable it follows by definition that $T$ is separable.