Finite Space is Second-Countable
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Theorem
Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ is a second-countable space.
Proof
Let $T = \struct {S, \tau}$ be a topological space where $S$ is finite.
As $S$ is finite, then so is its power set $\powerset S$.
So is $\tau \subseteq \powerset S$.
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.
$\blacksquare$