Prove that in a metric space $X$, a subset $S \subset X$ is relatively sequentially compact if and only if its closure $\overline{S}$ is sequentially compact.
The terms relatively sequentially compact and sequentially compact are defined as follows for a general topological space $X$:
A subset $S \subset X$ is sequentially compact if each sequence in $S$ has a convergent subsequence in $X$ with a limit in $S$.
A subset $S \subset X$ is relatively sequentially compact if each sequence in $S$ has a convergent subsequence in $X$.
My attempt for proving the first implication is the following:
Suppose $S$ is relatively sequentially compact. Suppose $(x_n)_n \subset \overline{S}$ with $x_n \in S \subset \overline{S}$, for all $n$. Since $S$ is relatively sequentially compact and $\overline{S}$ is closed, the sequence $(x_n)_n$ has a convergent subsequence with limit in $\overline{S}$.
I have difficulties with completing the proof of this first implication. I think the converse implication is clear:
Suppose $\overline{S}$ is sequentially compact. Then, in particular, any sequence $(x_n)_n \subset S$ has a converging subsequence with limit in $\overline{S} \subset X$.
Any help, solutions or comments are highly appreciated.