Every complete, countable metric space has a discrete, dense subset.
Solution 1:
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 $U$.)