I'm currently trying to get my head around a proof in Lipschutz [1]. Solved exercise 21 of Chapter 11 (page 164) asks for a proof of the following statement: Let $A$ be a countably compact subset of a metric space $X$. Then $A$ is also sequentially compact.
I can follow most of the proof. But there is one step which I'm unable to understand: since $X$ is a metric space, we can choose a subsequence of any sequence in $A$, which converges to the point $p$ in $A$.
I can see that the point $p$ is already known to be an accumulation point of the set of points in the sequence we're considering, since $A$ is countably compact by hypothesis. But I don't see how convergence to that point follows from $X$ being a metric space. I feel I'm probably missing something obvious about metric spaces, because I don't see it covered explicitly anywhere previously in the book.
[1] "General Topology", by Seymour Lipschutz (ISBN 0070379882)
Update:
More details...
A sequence $<a_1, a_2, ...>$ in $A$ is considered, in the two separate cases, where the set $B = \{a_1, a_2, ...\}$ is either finite or infinite. I won't reproduce the finite case here (but I can do so, if requested).
In the case where $B$ is infinite, it contains an accumulation point in $A$ (because $A$ is countably compact by hypothesis). Because $X$ is a metric space, we can choose a subsequence $< a_{i_1}, a_{i_2}, ... >$ of the original sequence which converges to the point $p$ in $A$ (i.e., $A$ is sequentially compact).
That's basically the solution provided in the book. I tweaked the wording a bit, because I'm paranoid and I don't know if McGraw-Hill is gratuitously litigious.
"contains" is the wrong word. $B$ has an accumulation point $p$ in $A$, but we don't generally have $p\in B$, so $B$ doesn't (generally) contain an accumulation point.
Language criticism done, the crucial fact is that a metric space is first countable, each point has a countable neighbourhood basis (for example the balls $B_{2^{-n}}(x)$ around $x$ with radius $2^{-n}, \; n \in \mathbb{N}$ form a countable neighbourhood basis of $x$).
And then we can construct a subsequence converging to $p$ as follows:
Since $p$ is an accumulation point of $B$, there are infinitely many points in $B\cap B_1(p)$. Choose an $n_1$ such that $a_{n-1} \in B_1(p)$.
Since $p$ is an accumulation point of $B$, there are infinitely many points in $B\cap B_{1/2}(p)$. Thus there is an $n_2 > n_1$ such that $a_{n_2}\in B_{1/2}(p)$.
Having found $n_1 < n_2 < \dotsc < n_k$, since $p$ is an accumulation point of $B$, there are infinitely many points in $B\cap B_{2^{-(k+1)}}(p)$. Thus there is an $n_{k+1} > n_k$ with $a_{n_{k+1}} \in B_{2^{-(k+1)}}(p)$.
Inductively, you obtain a subsequence $(a_{n_k})$ with $a_{n_k} \in B_{2^{-k}}(p)$ for all $k$, and that subsequence then converges to $p$.