Indiscrete Topology is not Metrizable

Theorem
Let $S$ be a set with more than one element.

The indiscrete topology on $S$ is not metrizable.

Proof
In order to be metrizable, there needs to be a metric $d$ on $S$, so that $\left({S, d}\right)$ is a metric space.

As $S$ has more than one element, $\exists x, y \in S: x \ne y$.

Then $\epsilon = d \left({x, y}\right) > 0$.

So the $\epsilon$-neighborhood $N_{\epsilon / 2} \left({x}\right)$ is $d$-open.

Hence $N_{\epsilon / 2} \left({x}\right)$ is in the topology which $d$ induces.

But $x \in N_{\epsilon / 2} \left({x}\right)$ while $y \notin N_{\epsilon / 2} \left({x}\right)$.

Thus $N_{\epsilon / 2} \left({x}\right) \ne \varnothing$ and $N_{\epsilon / 2} \left({x}\right) \ne S$.

So the topology induced by $d$ is not the indiscrete topology.