TITLE: $(X,\mathscr T)$ is compact $\iff$ every infinite subset of $X$ has a complete limit point in $X$.
QUESTION [12 upvotes]: Let $(X,\mathscr T)$ be a topological space. Given $A\subseteq X$, we say that $x$ is a complete limit point of $A$ if for every nbhd $N$ of $x$, $|N\cap A|=|A|$. I want to prove 

Suppose $(X,\mathscr T)$ is compact. Then every infinite subset $A$ of $X$ has a complete limit point in $X$.

PROOF Assume the contrary. Then there exists an infinite subset $A$ of $X$ such that for any $x\in X$, there is a nbhd $N_x$ such that $|N_x\cap A|<|A|$. The collection $\{N_x:x\in X\}$ covers $X$. By compactness, there exists a finite subcover $\{N_{x_1},\ldots, N_{x_n}\}$. Then $A=\bigcup_{i=1}^n A\cap N_{x_i}$. But this is impossible, since $A$ is infinite, and $\bigcup_{i=1}^n A\cap N_{x_i}$ is a finite union of sets strictly smaller than $A$.
Is this correct? The problem is in "Introductory Real Analysis" by Kolmogorov.
Also, is a proof of the converse too complicated? Kolmogorov gives a reference as "P.S. Alexandroff, Einführung in die Mengenlehre und die Theorie der Reellen Funktionen (1956), pp. 250-251; J.L. Kelley, General Topology (1955), pp. 163-164" but I have no idea what this means. I am honestly clueless on how to approach it. Could you provide hints? I wouldn't mind a complete answer, if you think it is requires something too advanced.

REPLY [8 votes]: Your proof is correct.  For the converse, suppose $X$ is not compact.  Let $\mathcal{O}$ be an open cover of $X$ of minimal cardinality [such that no finite subcover exists], and let $\{O_\alpha\}_{\alpha < |\mathcal{O}|}$ be an enumeration of $\mathcal{O}$.  Construct a set $A$ as follows:

let $x_\alpha \in X$ be any element not covered by any of the $O_\beta$ for $\beta < \alpha$.  Such $x_\alpha$ exists or else $\{O_\beta\}_{\beta < \alpha}$ would be an open cover, contradicting the minimality of $|\mathcal{O}|$.
let $A = \{x_\alpha : \alpha < |\mathcal{O}|\}$

We'll show that $A$ has no complete limit point.  Indeed, for any $x \in X$ there is some $\alpha$ such that $O_\alpha$ is a neighbourhood of $x$, but $O_\alpha$ avoids all the $x_\beta$ for $\beta > \alpha$, and so:
$$\left |A \cap O_\alpha\right | \leq \left|\{x_\gamma : \gamma \leq \alpha\}\right| = |\alpha| < |\mathcal{O}| = |A|$$