Show that a subset $Y$ of metric space $X$ is separable if there exists a sequence of points in $X$ whose closure contains $Y$
Solution 1:
Fact: every subspace of a separable metrizable space is separable (see here).
Let $C$ be the closure of a sequence in $X$ and assume that $Y\subset C$. Since $C$ is the closure of a countable set, it is separable. Then, $Y$ is separable, being a subset of $C$.
Note: metrizability is necessary for the fact to hold. Counterexample: the anti-diagonal in the Sorgenfrey plane is a non-separable subspace of a separable space.