Indiscrete Space is Separable

Theorem
Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space such that $S$ has more than one element.

Then $T$ is separable.

Proof
By definition, $T$ is separable there exists a countable subset of $S$ which is everywhere dense in $T$.

Let $x \in T$.

Then $\left\{{x}\right\} \subseteq T$ and $\left\{{x}\right\}$ is (trivially) countable.

From Subset of Indiscrete Space is Everywhere Dense we have that $\left\{{x}\right\}$ is everywhere dense.

Hence the result by definition of separable space.