I've seen lots of casual claims that Riemannian manifolds (even without assuming second-countability) are metrizable. In the path-connected case, we can use arc-length to create a distance function. But how can one prove this without assuming path-connectedness (and second-countability)?
(Motivated by Problem 2-D of Milnor-Stasheff.)
On
Assume the manifold $M$ is connected. If there exists a piecewise-smooth path from $p$ to $q$ in $M$, then there exists such a path between $p$ and $q'$ for any $q'$ in a small closed neighborhood of $q$, since $M$ is locally path-connected. The set of such $q$ is thus open and closed in $M$, and so must be $M$ itself. It follows that $M$ is path-connected, and you can define the metric on $M$ by arc-length.
By the Smirnov metrization theorem, a space that is locally metrizable and paracompact is metrizable. So as each path component of a Riemannian manifold is metrizable as you explain, local metrizability clearly follows. Paracompactness is often required in the definition of a manifold (or that it is second countable, which implies paracompactness), which is why this claim is often made without additional hypotheses.
But even in the non-second countable case, as each path component of a Riemannian manifold is metrizable, it must be paracompact, again by Smirnov. An arbitrary disjoint union of paracompact spaces is paracompact, a quick proof is here, so a Riemannian manifold must be paracompact.
See also this related question.