Separable Space need not be First-Countable

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is separable.

Then $T$ does not necessarily have to be first-countable.

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

We have that a Finite Complement Topology is Separable.

But we also have that an Uncountable Finite Complement Space is not First-Countable.

Hence the result, by Proof by Counterexample.