Baby Rudin 2.26 Infinite subsets with limit points implies compactness Having some trouble with this question.

Let $X$ be a metric space in which every infinite subset has a limit
  point. Prove that $X$ is compact.  Hint: By Exercises 23 and 24, $X$
  has a countable base. It follows that every open cover of $X$ has a
  countable subcover ${G_n}$, $n = 1, 2, 3, ....$ If no finite
  subcollection of ${G_n}$ covers $X$, then complement $F_n$ of $G_1 \cup \dots \cup G_n$ is nonempty for each $n$, but $\bigcap F_n$ is empty. If $E$
  is a set which contains a point from each $F_n$, consider a limit
  point of $E$, and obtain a contradiction.

I cannot justify the phrase "It follows that every open cover of $X$ has a countable subcover ${G_n}$, $n = 1, 2, 3, ....$", am I overlooking something simple? Is there a simple map between the countable base and any open subcover?
 A: Let $\mathscr{B}$ be a countable base for $X$, and let $\mathscr{U}$ be an open cover of $X$. For each $x\in X$ there is a $U_x\in\mathscr{U}$ such that $x\in U_x$, and there is a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq U_x$. Let $\mathscr{B}_0=\{B_x:x\in X\}$; clearly $\mathscr{B}_0$ is a countable cover of $X$. The construction of $\mathscr{B}_0$ ensures that for each $B\in\mathscr{B}_0$ there is a $U_B\in\mathscr{U}$ such that $B\subseteq U_B$; let $\mathscr{V}=\{U_B:B\in\mathscr{B}_0\}$. I claim that $\mathscr{V}$ is a countable subcover of $\mathscr{U}$.


*

*Clearly $\mathscr{V}\subseteq\mathscr{U}$.  

*$\mathscr{V}$ is indexed by the countable set $\mathscr{B}_0$, so $\mathscr{V}$ is countable.  

*Let $x\in X$; by construction $x\in B_x\in\mathscr{B}_0$, so $x\in B_x\subseteq U_{B_x}\in\mathscr{V}$, so $\mathscr{V}$ covers $X$.

A: I do not know if this is what Rudin had in mind, but if we let $\mathcal{Y}$ be an open cover of $X$. Then since $X$ has a countable base $\mathcal{B}$, each open set in $\mathcal{Y}$ can be written as a union of (at most) countably many elements in $\mathcal{B}$. So you can choose a countable subcover of $\mathcal{Y}$ by choosing open sets of the form $\mathcal{Y}_{\alpha}:B_{\alpha}\in \mathcal{Y}_{\alpha}$. Here $\alpha$ is an index set such that $\bigcup B_{\alpha}=X$. The choice of $\mathcal{\alpha}$ may not be unique and some $\mathcal{Y}_{\alpha}$ might even coincide or empty, but after adjusting indexes we can get a countable subcover as desired. This construction seems identical with the comment above, though.  
A: Your question, in clear words and proper terminology, is how to show that every second-countable topological space (or even just metric space) is Lindelöf.
To see that note, that if $\cal U$ is an open cover and $\mathcal B=\{V_n\mid n\in\Bbb N\}$ is a countable basis, then taking $\mathcal V_n=\{U\in\mathcal U\mid V_n\subseteq U\}$ is a countable family. It might be that some $V_n$ are empty, in which case we omit them. But we are left with a family of at most $\aleph_0$ non-empty sets. Pick $U_n\in\mathcal V_n$.
I claim that $\{U_n\mid n\in\Bbb N\}$ is an open cover. Those are all open sets, so we just need to show it's a cover. Let $x\in X$ be an arbitrary point, then for some $n\in\Bbb N$ we have that $x\in V_n$, and for some $U\in\mathcal U$ we have that $x\in U$. Therefore $U\cap V_n$ is a non-empty open set and so there is some $m$ such that $x\in V_m\subseteq U\cap V_n$.
Therefore $x\in U_m$ by the definition of $\mathcal V_m$, as wanted.
