= 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 \b[a subset of a <metric space> is a <closed set> exactly when it is sequentially closed].
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