Let X be a metric space in which every infinite subset has a limit point. Prove that X is compact. Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is compact.
The following is my proof I'd like to know if it is correct.
Proof:
I will use the fact that if $X$ is a metric space in which every infinite subset has a limit point, then $X$ has a countable base, so every open cover of $X$ has a countable subcover $\{G_n\}$. To lead contradiction, suppose that there is no finite subcollection of $\{G_n\}$ which covers $X$. Then the complement $F_n$ of $G_1 \cup \dots \cup G_n$ is nonempty for each $n$, while clearly $\bigcap F_n$ is empty. So we can choose $x_1$ that doesn't belong to $G_1$ and $x_2$ distinct from $x_1$ that doesn't belong to $G_1 \cup G_2$ and recursively, get a sequence of distinct points $\{x_n\}$ such that each $x_n$ is from $F_n$. Let's call this set $E$ then $E$ is infinite (since if it is finite then $E=\{x_1,\dots, x_k\}$. Then each $x_i$ is covered by some $G_m$, we can take a union of at most $k$ such covers and since it is only a finite union, by assumption, there exists another point outside the union that belongs to $E$.) 
Now by hypothesis, $E$ has a limit point $z$. Then, $z$ belongs to some open cover $G_m$ and so there is an open ball $B(z;d)\subset G_m$. But then, by our construction, all $x_n$ for which $n \ge m$, do not belong to $B(z;d)$. So there is an open ball containing $z$ which doesn't contain infinitely many points of $E$, which contradicts the assumption that it is a limit point. Thus, by way of contradiction, there must be a finite subcover. QED.
This is my proof but I have a couple points that I'm not sure of. First, how do I guarantee that I can select all distinct points of $x_n$ by the above process? Also, is my proof that $E$ must be an infinite set correct? Is there anything incorrect about the rest of the proof? Thanks.
 A: If no finite subcollection of $\{G_{n}\}$ covers X, then $F_{1} = G_{1}^{c}\neq\emptyset$, so choose $x_{1}\in F_{1}$. Having chosen $x_{1}, . . .,x_{j} \in X$, choose $x_{j+1} \in X$, if possible, so that $x_{j+1} \in F_{j+1} \setminus \{x_{1}, ...,x_{j}\}$. If I prove that this process must not stop after a finite number of steps, then you will get distinct $x_{n}'s$ and the set E will be infinite.
suppose this process stops after a finite number of steps, say k, then $F_{k+1}\setminus \{x_{1},...,x_{k}\}=\emptyset$. Then $G_{1}\cup ....\cup G_{k+1}\cup\{x_{1},...,x_{k}\}=X$. Since $\{G_{n}\}$ covers X, there exists $G_{m_{i}}$ such that $x_{i}\in G_{m_{i}}$ for $i = 1,...,k$. Now $G_{1}\cup ...\cup G_{k+1}\cup G_{m_{1}}\cup ...\cup G_{m{k}} = X$, a contradiction.
A: Here's part of my proof:
Suppose that E contains a point of each $F_{n}$. If E is infinite, then it has a limit point $p$. Notice that if we had $p\in G_{n}$ for some $n$, then there would be a neighborhood with no points of $F_{n+1}$. Then $p\notin G_{n}\forall n$, that is, $p\in \bigcap F_{n}$. 
Hint: Consider the set $\bigcup_{n}F_{n}$. What can be said about the relationship between $F_{n}$ and $F_{n+1}$? After answering that, consider the set $\bigcap_{n}F_{n}$ to get a contradiction.
A: Any sequence will then have a limit point, so a subsequence of that sequence will converge to the limit point.
