Indiscrete Space is T3

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

Then $T$ is a $T_3$ space.

Proof
Let $F \subseteq X$ be a closed set in $S$, and $y \in X$ such that $y \notin F$.

The only way this can happen is if $F = \varnothing$.

So there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.

That is, $U = \varnothing$ and $V = S$.

Hence (trivially) the result.