Limit point Compactness does not imply compactness counter-example

Solution 1:

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.