Showing that a set X with a trivial topology is not metrizable

You're almost there.

For a space to have a metric, you must be able to distinguish any two points, that is: $d(x,y)=0$ if and only if $x=y$. But the indiscrete topology has way too few open sets for this to be possible, i.e. there cannot be any $\epsilon$-balls separating $x$ from $y$.

For the discrete topology, here's a hint: the discrete metric.

(alternatively: every metric space is Hausdorff)

That $X$ has more than one element is implicit in the above answers. If $X$ has exactly one element the discrete and indiscrete topologies coincide and are metrizable.

the discrete topology is metrizable, $d(x,y)=1$ for all $x\neq y$. the indiscrete (if$|X|>1$) is not metrizable since points will not be closed.