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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.

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    There are 2 notions of limit point here: $x$ is a limit point of $A$ iff every open ball around $x$ intersects $A$ in a point unequal to $x$. Also $x$ is called an accumulation point of $A$ if every ball around $x$ intersects $A$ in an infinite set. Both are often called limit points. They are equivalent for metric spaces, or more generally $T_1$ spaces. The proof shows $x$ is not an accumulation point, as the ball contains at most finitely many $x_n$, namely,thsoe with index $ – 2017-02-17
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    The wording of the proof is a bit poor, but $E$ is _defined_ as the set $\{x_m\}$, so of course they should have the same set of limit points. – 2017-02-18

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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 $E$.

Response:

Note: There's a typo in the above Question, that is, the last usage of $x_m$ in the proof must be replaced by $\{x_m\}_{m \geq n}$. It's the sequence that that proof is referring to.

The facts that $B(z,\epsilon)$ sits inside $G_n$ and $x_m \notin G_n, \forall~ m \geq n$ tells us that $z$ cannot be a limit point of the sequence $\{x_m\}_{m \geq n} $.

Now mimicking the steps provided by @user417252. The only elements of $E$ that $B(z,\epsilon)$ can possibly contain are among $x_1, \cdots, x_n$ by the above argument. Let $\delta$ be the minimum of $\epsilon$ and the distances $d(z,x_i)$($d$ is a metric of the metric space in which we are working) for those values of $i = 1, \cdots, n$ such that $z \ne x_i$. Then $E \cap (B(z,\delta) - \{z\}) = \varnothing$, hence $z$ cannot be a limit point of $E$.

The following is unnecessary from the previous edit: Now $\{x_m\}_{m \geq n}$ being the subsequence of $E$, tells us that $z$ is not a limit point of $E$. (Exercise: Sequence $\{a_n\}$ converges to $a$ if and only if every subsequence of $\{a_n\}$ converges to $a$.)

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    You're confusing the limit of a sequence with [limit points](https://en.wikipedia.org/wiki/Limit_point). As an example of the distinction, the sequence $(0,1,0,1,\dotsc)$ in $\mathbb R$ has no limit, but it has _two_ limit points $0$ and $1$. In particular, limit points of a sequence need not be a limit point of every subsequence (in the previous example, $1$ is not a limit point of $(0,0,0,\dotsc)$). So the implication in the last paragraph is unjustified and the exercise irrelevant. – 2017-02-18
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    Thanks @epimorphic for the comment. Have updated the answer. Got mixed up *limit* and *limit points* and used chapter 3's results in the previous edit. – 2017-02-18
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IIRC Rudin defines a limit point of $E$ as a point $x$ such that $B(x,r) \cap (E \setminus \{x\}) \neq \emptyset$ for every $r>0$. He then has a lemma of sorts that shows that :

If $x$ is a limit point of $E$, then for each $r>0$, $B(x,r) \cap E$ is infinite. This uses the fact that all finite subsets of a metric space are closed.

Then to the proof: suppose that $G_n, n \in \mathbb{N}$ is a countable open cover without a finite subcover. The no finite subcover condition allows us to take a point $x_1 \notin G_1$ ,$x_2 \notin G_1 \cup G_2$, and in general $x_n \notin \cup_{i=1}^n G_i$. Then define $E = \{x_n: n \in \mathbb{N}\}$ which is an infinite set. By assumption $E$ has a limit point $p$. This $p$ is covered, so for some $m$, $ \in G_m$, and as $G_m$ is open, for some $r>0$, $B(p,r) \subset G_m$. Now, for all $n>m$, by construction $x_n \notin \cup_{i=1}^n G_i$, so in particular $x_n \notin G_m$(as $n > m$), so $x_n \notin B(p,r)$.

So $B(p,r) \cap E \subset \{x_n: n \le m\}$, as all $x_n$ with larger index have been ruled out above. So $B(p,r) \cap E$ is finite. So by the first paragraph's equivalence, $p$ cannot be a limit point of $E$. Contradiction.

So $X$ is countably compact: every countable open cover has a finite subcover. As you already know that $X$ has a countable base we can thin out any open cover of $X$ to a countable one first, and a finite one now, so compactness has been shown.

It's important to work with the infinite intersection variant of limit points here.

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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 $m > n$.

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    The set $E$ is infinite in this context and thus the $a$ in your answer won't exist. – 2017-02-17
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    @stud_iisc The distances to the remaining points of $E$ are all larger than $\epsilon$ by the definition of $\epsilon$ (and in any case user417252 made it explicit that the minimum is over $i$ between $1$ and $n$). – 2017-02-18
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    Thank you for making it clear OP and @epimorphic. – 2017-02-18