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 $\struct {S, d}$ is a metric space.

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

Then:
 * $\epsilon = \map d {x, y} > 0$

So the open $\epsilon$-ball $\map {B_{\epsilon / 2} } x$ is $d$-open.

Hence $\map {B_{\epsilon / 2} } x$ is in the topology which $d$ induces.

But $x \in \map {B_{\epsilon / 2} } x$ while $y \notin \map {B_{\epsilon / 2} } x$.

Thus $\map {B_{\epsilon / 2} } x \ne \O$ and $\map {B_{\epsilon / 2} } x \ne S$.

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