Compactness of closure Let $A$ be a subset of a metric space $X$.
Assume that each sequence in $A$ has a convergent subsequence with limit in the closure $\overline{A}$.
Does this imply that $\overline{A}$ is compact? I read it on wikipedia but I couldn't find a proof of it anywhere. 

Yes, that does imply that $\overline{A}$ is compact. For metric spaces, compactness is equivalent to sequential compactness, so let's consider an arbitrary sequence $(x_n)$ in $\overline{A}$. For each $n$, choose an $y_n \in A$ with $d(y_n,x_n) < 2^{-n}$. By the premise, there is an $x \in \overline{A}$ and a subsequence $(y_{n_k})$ with $\lim\limits_{k\to\infty} y_{n_k} = x$. Then we also have $\lim\limits_{k\to\infty} x_{n_k} = x$ (since $d(x,x_{n_k}) \leqslant d(x,y_{n_k}) + d(y_{n_k},x_{n_k}) < d(x,y_{n_k}) + 2^{-n_k}$), so $\overline{A}$ is sequentially compact.
