Complement Finite space is separable

Question

Sow that $(X, )$ with the complement finite topology (the closed sets are the finite subsets of $X$) is separable. If $X$ is countable it's easy but I'm interested in the non-countable infinite case.

Answer 1

Pick any countably infinite subset $D$ of $X$. We claim that this set is dense in $X$. Let $O\in\tau$ be open and non-empty. Then $O^c$ is finite, hence there exists $x\in D\cap O$.

As this holds for arbitrary non-empty $O\in\tau$, the set $D$ is dense.