 Solution

If $A$ is closed and $a_n\in A$ converges to $x\in X$, then $x$ cannot lie in the open complement: an open ball about such an $x$ would eventually contain all $a_n$ while missing $A$. Thus $x\in A$.

Conversely, suppose $A$ contains every limit of a sequence in $A$. If $x\in \overline{A}$, every ball $B(x,1/n)$ meets $A$; choose $a_n$ in that intersection. Then $a_n\to x$, so $x\in A$. Hence $\overline{A}=A$, proving that a subset of a metric space is a closed set exactly when it is sequentially closed.