Finite Complement Topology is Separable

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

Then $T$ is a separable space.

Proof
Let $H$ be a countably infinite subset of $S$.

From Closure of Infinite Subset of Finite Complement Space, the closure of $H$ is $S$.

So by definition $H$ is everywhere dense in $T$.

Hence the result by definition of separable space.