Indiscrete Space is Second-Countable

Theorem
Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.

Then $T$ is a second-countable space.

Proof
The only basis for $T$ is $\left\{{S}\right\}$ which is trivially countable.

Hence the result.