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.