Why does countable compactness imply compactness on metric spaces? By "$E$ is countably compact", I mean that every countable open cover of $E$ has a finite subcover.  By "$E$ is compact", I mean that every open cover of $E$ has a finite subcover. Let $M$ be a metric space.  Prove that if a subset $E$ of $M$ is countably compact, then it is compact.
I am looking for a proof that just uses the machinery of metric space topology, and doesn't appeal to general topological spaces, if this is possible.
 A: This is tied in to a standard set of compactness results for metric spaces.  Here the usual sequence of equivalent properties is as follows:


*

*Countable compactness

*Limit point compactness (every infinite set has a limit point)

*Sequential compactness (every sequence has a convergent subsequence)

*Compactness


These are all equivalent for metric spaces.  I'm afraid I don't know off the top of my head a more direct proof than proving the chain of implications one at a time.  I checked in Munkres, which proves that the latter three are equivalent (which should be in any standard text), and the only thing remaining to show is the implication (countable compactness) $\implies$ (limit point compactness), which uses the following modified argument.
You must prove that every set $A$ without limit points is finite.  We can assume that $A$ is countable without loss of generality (If every countably infinite set has a limit point, then all infinite sets have limit points).  Now for each element $a\in A$, there is a neighborhood $U_a$ of $a$ containing no other points of $A$ (since $a$ is not a limit point of $A$).  Since $A$ contains all of its limit points tautologically, $A$ must be closed, which in turn implies that $A$ is countably compact.  Therefore the cover $\{U_a\}$ has a finite subcover, implying that $A$ is finite.
