Does a separable metric space have a countable basis? I want to prove that if $X$ is a metric space and has a dense countable subset, then it has a countable basis. I know that every metric space is first countable, but I can't continue. Thanks for your help.

Well, take a dense subset of $D\subset X$ and then the metric guarantees you a countable neighbourhood basis at every point of $D$. A countable union of countable sets is again countable. Check that this is enough.
