Let $((E_i,\mathcal O_i))_{i\in I}$ be an at most countable family of topological spaces, each containing at least two distinct elements. Let $E=\prod_{i\in I}E_i$ and $\mathcal O$ be the product topology on $E$.
It is well known that if each $E_i$ is metrisable, then $E$ is metrisable.
Is the converse true? That is if $E$ is metrisable, is $E_i$ metrisable for all $i\in I$?
Yes, because each $E_j$ is homeomorphic to a subspace of $\prod_{i\in I}E_i$. And a subspace of a metrizable space is metrizable.