TITLE: Every complete, countable metric space has a discrete, dense subset.
QUESTION [15 upvotes]: Given a complete, countable metric space, say $X$, I'd like to show it has a discrete, dense subset. This seems like an application of the Baire Category Theorem, but that doesn't seem to go anywhere. Any help would be appreciated.

REPLY [11 votes]: Consider the collection $I$ of all isolated points of $X$.  (By the Baire Category Theorem $I$ is nonempty, but that is somewhat immaterial for the moment.)  Note that $I$ is then a discrete subspace of $X$.  If $I$ were not dense, then $U = X \setminus \overline{I}$ is a nonempty (open) set without isolated points.  From here we can construct in the usual manner a Cantor set as a subset of $X$, contradicting that $X$ is countable!  (The construction goes as in the linked answer, just ensuring that the $x_\sigma$ are chosen from $U$.)