We shall show that if $X$ is a $T_1$ space, then it is countably compact if and only if every infinite open cover has a proper subcover. The key idea is to proceed by contrapositive.
($\Rightarrow$ Needs $T_1$) Suppose that $ \ \mathcal{U} \ $ is an infinite open cover of $X$ with no proper subcover. Then for each $U \in \mathcal{U}$, there is a point $p_U \in U$ that doesn't belong to any other member of $ \ \mathcal{U} \ $. The set $A = \{p_U : U \in \mathcal{U} \}$ is infinite and doesn't have a limit point. Indeed, if $x \in X$, then there is $U \in \mathcal{U}$ containing $x$ and $ U \cap A = \{ p_U \}$. If $x = p_U$, then it isn't a limit point of $A$. If $x \ne p_U$, then, using $T_1$, $U- \{ p_U \}$ is open and $(U- \{ p_U \}) \cap A =\emptyset$. So $x$ isn't a limit point of $A$.
George Lowther gave an example showing that the $T_1$ hypothesis is fundamental. Since there are many comments, I'll reproduce it here:
"Consider the example of the real numbers where the open sets are unions of intervals $[n,a)$ for integer $n$ and real $a>n$. This is $T_0$ but not $T_1$. It is also countably compact, but the infinite cover $\mathcal{U}=\{[n,n+1)\colon n\in\mathbb{Z}\}$ has no proper subcovers. So, $T_1$ is needed."
A direct approach to prove this implication was suggested by Carl Mummert. Let $ \ \mathcal{U} \ $ be an infinite open cover of $X$ and $ \ \mathcal{U}_0 $ be a countably infinite subset of $ \ \mathcal{U} $. Now consider $ \ \mathcal{U}_1 = \mathcal{U} - \mathcal{U}_0$ and let $V \ $ be the union of all sets in $ \ \mathcal{U}_1$. Then $ \ \mathcal{U}_0 \cup \{ V \ \}$ is a countable open cover of $X$ and follows from $(1)$ that it has a finite subcover $U_1, \ldots, U_n$. Now the set consisting of all $U_i$, with possible exception of $V$, adjoined with the sets $ U \in \mathcal{U}_1$ is a proper subcover of $ \ \mathcal{U}$.
($\Leftarrow$ Doesn't need $T_1$) Now, suppose that $X$ isn't countably compact. Then there is an infinite set $A$ with no limit points. Thus $A$ is closed. By the definition of limit point, for each $x \in A$, there is an open set $U_x$ containing $x$ such that $ U_x \cap A = \{ x \} $. Consider $ \ \mathcal{U} \ $ the set of all $U_x$ plus, if necessary, $X-A$. Then $ \ \mathcal{U} \ $ is an infinite cover of $X$ with no proper subcovers.
P.S. Thanks Carl Mummert, George Lowther and Mark for your efforts to clarify and improve this answer.
You are correct: that space is not countably compact. Indeed, a space that is countable and countably compact is automatically compact, since every open cover certainly has a countable subcover.
One of the simpler examples of a space that is countably compact but not compact is $\omega_1$, the space of countable ordinal numbers, with the order topology. If for each $\alpha<\omega_1$ we let $U_\alpha=[0,\alpha)$, then $\{U_\alpha:\alpha<\omega_1\}$ is an open cover of $\omega_1$ with no finite subcover. However, every infinite subset of $\omega_1$ has a limit point in $\omega_1$, so $\omega_1$ contains no infinite closed discrete set and is therefore countably compact.
To see this, suppose that $A\subseteq\omega_1$ is infinite. Let $\alpha_0=\min A$, and for each $n\in\omega$ let $\alpha_{n+1}=\min\{\alpha\in A:\alpha>\alpha_n\}$. Then $\langle\alpha_n:n\in\omega\rangle$ is a strictly increasing sequence in $A$. Let $\alpha=\sup_{n\in\omega}\alpha_n$; then $\alpha<\omega_1$, and $\alpha$ is easily seen to be a limit point of $A$.
Best Answer
This is tied in to a standard set of compactness results for metric spaces. Here the usual sequence of equivalent properties is as follows:
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.