Prove a metric space in which every infinite subset has a limit point is compact.(Question about one particular proof) This is an exercise from "Baby Rudin" and I don't understand one step of a particular proof based on the textbook's hint.
$\mathbf{Statement:}$Let X be a metric space in which every infinite subset has a limit point. Prove that X is compact.
$\mathbf{Hint:}$ 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 
 $G_1 \cup.. \cup G_n$ is nonempty for each n, but $\cap F_n$ 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.
$\mathbf{Proof:}$ Following the hint, we consider a set consisting of one point from the complement of each finite union, e.t.,$x_n\notin G_1∪⋯∪G_n$. Then E cannot be finite. By hypothesis E must have a limit point z, which must belong to some set $G_n$；and since $G_n$ is open, there is a $\epsilon >0 $ such that $B(z,\epsilon) \subset G_n$ . But then $B(z,\epsilon)$ cannot contain $x_m$ if $m \gt n$, and so z cannot be a limit point of ${x_m}$. We have now reached a contradiction.
$\mathbf{Question:}$ I don't understand the causal relation between "z cannot be a limit point of ${x_m}$." and that z is not a limit point of E.

The only elements of $E$ that $B(z,\epsilon)$ can possibly contain are among $x_1, \dots, x_n$. Let $a$ be the minimum of $\epsilon$ and the distances $d(z,x_i)$ for those values of $i = 1, \dots, n$ such that $z \ne x_i$. Then $E \cap (B(z,a) - \{z\}) = \varnothing$, hence $z$ cannot be a limit point of $E$. 
Edit: In response to stud_iisc below, the proof says that $E$ consists of all elements $x_m$, and that $x_m \not\in B(z,\epsilon)$ for $m > n$.
