I think that I understand why compactness implies limit point compactness:
Suppose $A \subseteq X$ has no limit points. Then $A^{\prime} \subseteq A$. Thus, $A$ is closed. Then for all $a \in A$, there exists some neighborhood $U_a$ of $A$ s.t. $U_a \cap A= \{a\}$. But then since $X$ is compact and $A \cup (X-A)=X$, it is an open covering. [$X-A$ has to be open, since $A$ is closed]. Then by compactness, there is a finite open covering and since $(X-A) \cap A = \emptyset$, It is clear that $A$ is finite. The result follows
However, the converse of the statement is supposedly incorrect. This definitely seems weird, particularly in $R$ (equipped with usual topology.)
Can somebody provide me with a counter-example?
Best Answer
Your proof is not correct as it stands, as $U_a$ is not equal to $\{a\}$ but we know that $U_a \cap A \subseteq \{a\}$, in general.
We also don't really need the fact that $A$ is closed (though this is true). Pick for every $x \in X$ some open $U_x$ that contains $x$ and such that $U_x \cap A \subseteq \{x\}$. This is, by definition almost, an open cover of $X$. Compactness tells us that there is a finite subset $F$ of $X$ such that $X = \cup_{x \in F} U_x$. But then $A = X \cap A = A \cap (\cup_{x \in F} U_x) = \cup_{x \in F} (U_x \cap A) \subseteq F$, by how the $U_x$ were chosen, and this contradicts that $A$ is infinite.
A countably compact space that is not compact is the first uncountable ordinal, $\omega_1$, in the order topology. Or $\{0,1\}^\mathbb{R} \setminus \{\underline{0}\}$ and many more.
You are right that for metric spaces the reverse does hold, but this is atypical.