Countable Complement Space is not Separable

Theorem
Let $T = \left({S, \tau}\right)$ be a countable complement topology on an uncountable set $S$.

Then $T$ is not a separable space.

Proof
Let $U$ be a countable subset of $S$.

By the definition of $T$, $U$ is closed.

As $U$ is countable but $S$ is uncountable:
 * $U \subsetneq S$

and so:
 * $U \ne S$

From Closed Set Equals its Closure:
 * $U^- = U$

where $U^-$ is the closure of $U$.

Thus:
 * $U^- \ne S$

So by definition, $U$ is not everywhere dense in $T$.

As $U$ is arbitrary, there is no countable set which is everywhere dense in $T$.

The result follows by definition of separable space.