Finite Space is Second-Countable

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

Then $T$ is a second-countable space.

Proof
Let $T = \left({S, \tau}\right)$ be a topological space where $S$ is finite.

As $S$ is finite, then so is its power set $\mathcal P \left({S}\right)$.

So is $\tau \subseteq \mathcal P \left({S}\right)$.

Now $\tau$ is a basis for itself, as every set in $\tau$ is a union of sets from $\tau$ (just one).

But $\tau$ is finite, and so countable.

The result follows from definition of a second-countable space.