Show that a subset $Y$ of metric space $X$ is separable if there exists a sequence of points in $X$ whose closure contains $Y$

Question

I'm trying to show that a subset $Y$ of metric space $X$ is separable if there exists a sequence of points in $X$ whose closure contains $Y$.

Attempt: I know that separable means that $Y$ must contain a countable dense subset. If we denote the sequence $\{x_n\}$ then it would seem that its closure should also be $\{x_n\}$, and potentially any limit point of $\{x_n\}$ in case the sequence converges, considering we may think of $\{x_n\}$ as the union of singelton sets $x_1\cup x_2 \cup x_3 \cup …$ and each singleton set has an open complement (as derived here) and thus the union of all the complements will be open and therefore the sequence is closed. Since the closure is the smallest closed set containing $\{x_n\}$ the closure would be $\{x_n\}$. But I would think that I somehow missunderstand what the closure of $\{x_n\}$ constitutes and therefore I'm unable to make progress.

Answer 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.