Compactness of a set in a metric space Let $X$ be a metric space, and $A\subset X$. Show that $\bar A$ is compact iff for every sequence in $A$ there exists a subsequence that converges to a point in $X$. I showed the "forward" direction, but am stuck showing the reverse.
Let $(x_n)\subset \bar A$ be a sequence. if $A$ is closed, then $A=\bar A$ and if a sequence in a closed set converges, it converges to an element of the set. But by assumption every sequence in $A$ has a convergent subsequence (to an element in $A$), and the result follows.
Thing is, what happens if $A$ is not closed? Choose a sequence in $\bar A$ such that every element of the sequence is not in $A$, i.e. $(x_n)\subset\partial A \setminus A$. What guarantees us that it has a convergent subsequence? Thanks!

Hint: if $(x_n)\subset \overline A$, we can find a sequence $(y_n) \subset A$ such that for all $n$, $d(x_n,y_n)\le 1/n$.
