Compactness and separability Just wanted to confirm I am on a right track. This is one of the problems from Rudin's "Principles of Mathematical Analysis". Any help/comment is HUGELY appreciated!
Given X is a metric space in which every infinite subset has a limit point. Prove that X is compact.
Assuming that I have already proven that X has a countable base. These are my proof steps:


*

*Since X has a countable base, every cover has a countable sub-cover.
The task is to prove that this countable cover is finite. By contradiction:

*Take one such countable sub-cover $\left\{G_n\right\}$

*Assume $\forall N\in \mathbb{N}.X\nsubseteq\bigcup_{n=1}^NG_n$

*Lets construct set: $\left\{V_n\right\}$ such that $V_1 = G_1$, $$V_n=\left\{x\in X:x\in \left[\left(\cup_{m=1}^{n-1} G_m\right)^c\bigcap G_{n}\right]\right\},n=2,3,...$$ Then $\forall n.V_n\subseteq G_n$; $V_n$ is open and $\forall n.\bigcup_{i=1}^nV_i =\bigcup_{i=1}^nG_i$. $\{V_n\}$ is also disjoint

*Lets construct an infinite set $$E = \left\{x_n\in X:x_n\in V_n\right\},n=1,2,...$$

*From the premise this set should have a limit point in X. Lets show that this is not true.

*Assume $\exists x\in X.x=lim\ E$ Then either $x\in E$ or $x\notin E$

*if  $x\in E$ then $x$ is an internal point is some $V_n$ and there exist an open ball around $x$ that is entirely in $V_n$ that doesn't contain any other members of $E$. Hence $x$ is not a limit point.

*If $x\notin E$ then $x$ is still and internal point in some $V_n$ and there exists an open ball around $x$ which is contained entirely in $V_n$. Also since X contains a countable base, $\forall p\in E. p\in V_n$ can be excluded from this open ball. Hence $x$ is not a limit point of E

*Steps 8 and 9 have shown, there is no limit point of $E$ in X. This is contradiction to assumption in step 3. Hence $\left\{G_n\right\}$ must be finite and X is compact.


Thanks,
Leon
 A: For $n>1$ the set $V_n$ need not be open: it’s the intersection of the open set $G_n$ with the closed set $X\setminus\bigcup_{k=1}^{n-1}G_k$. This means that your step $(8)$ can’t necessarily be carried out.
In fact you cannot hope to prove that the countable cover $\mathscr{G}=\{G_n:n\in\Bbb Z^+\}$ is actually finite, because it may not be: 
$$\left\{\left[0,\frac12\right)\right\}\cup\left\{\left(\frac1n,1\right]:n>1\right\}$$
is a countably infinite open cover of $[0,1]$. What you need to prove is that $\mathscr{G}$ has a finite subcover. You’ve actually got the right general idea for doing this; you just haven’t carried it out quite right. Here’s one way to do it.
For $n\in\Bbb Z^+$ let $U_n=\bigcup_{k=1}^nG_k$; we’d like to show that $U_n=X$ for some $n\in\Bbb Z^+$. Suppose not. Let $n_1=1$, and pick any $x_1\in U_{n_1}=G_1$. Since $U_{n_1}\ne X$, there is an $x_2\in X\setminus U_{n_1}$, and there is an $n_2>n_1$ such that $x_2\in U_{n_2}$. If $\ell\ge 3$, and we’ve already chosen $x_k$ and $n_k$ for $k<\ell$ so that $x_k\in U_{n_k}\setminus U_{n_{k-1}}$ for $k=2,\ldots,\ell-1$, we know that $U_{n_{\ell-1}}\ne X$, so we can choose $x_\ell\in X\setminus U_{n_{\ell-1}}$ and $n_\ell>n_{\ell-1}$ such that $x_\ell\in U_{n_\ell}$. In this way we recursively construct a set $D=\{x_k:k\in\Bbb Z^+\}$ and integers $n_k$ for $k\in\Bbb Z^+$ such that $x_k\in U_{n_k}$ for each $k\in\Bbb Z^+$, and $x_k\notin U_{n_{k-1}}$ for $k\ge 2$.
Now show that $D$ is an infinite, closed, discrete set in $X$, and you have your contradiction. For example, in place of your $(8)$ you can argue that $U_{n_k}$ is an open nbhd of $x_k$, and $x_k\ne x_\ell$ for $\ell<k$, so there is an $\epsilon_k>0$ such that $B(x_k,\epsilon_k)\subseteq U_{n_k}\setminus\{x_1,\ldots,x_{k-1}\}$, and therefore $D$ is discrete. You can modify your $(9)$ similarly. (Note that in $(9)$ you do not need the fact that $X$ has a countable base.)
