Help understanding why a complete, totally bounded metric space implies every infinite subset has a limit point I'm reading the following proof. Properties II and III are in my title, that a complete, totally bounded metric space implies every infinite subset has a limit point.

I have two questions near the end. Why does $d(x_n,x)\lt 2/n$? And secondly, why is $x$ a limit point of $A$? What other point in the neighborhood of $x$ is also in $A$? I don't get why they mention that $3/n\to 0$ as $n\to\infty$. How does that imply it's a limit point?
 A: There’s a slight error in the proof: the claim should be that $d(x_n,x) \le 2/n$. To see this, let $\epsilon$ be any positive real number; the sequence of $x_m$’s converges to $x$, so there is a positive integer $m > n$ such that $d(x_m,x) < \epsilon$. By the triangle inequality $$d(x_n,x) \le d(x_n,x_m)+d(x_m,x) < \frac2n + \epsilon.\tag{1}$$ Thus, $$d(x_n,x) < \frac2n + \epsilon$$ for every $\epsilon > 0$, and hence $d(x_n,x) \le \dfrac2n$.
This small error doesn’t affect the next step of the argument: if $y \in B(x_n,1/n)$, then $d(x_n,y) < 1/n$, so $$d(x,y) \le d(x,x_n)+d(x_n,y) \le \frac2n + d(x_n,y) < \frac2n + \frac1n = \frac3n,$$ $y \in B(x,3/n)$, and therefore $B(x_n,1/n) \subseteq B(x,3/n)$.
Now $B(x_n,1/n)\cap A$ is infinite for each $n$, and $B(x_n,1/n) \subseteq B(x,3/n)$, so $B(x,3/n)\cap A$ is infinite for each $n$. Since $3/n\to 0$ as $n\to\infty$, for any $\epsilon > 0$ there is an $n_\epsilon$ such that $3/n_\epsilon < \epsilon$. But then $B(x,\epsilon)\cap A \supseteq B(x,3/n_\epsilon)\cap A$, which is infinite. Thus, every nbhd of $x$ contains infinitely many points of $A$.
Added: To clarify, it is in fact true that $d(x_n,x)<2/n$ for each $n$; it just doesn’t follow directly from the fact that $d(x_m,x_n) < 2/m$ when $m<n$, as the weaker inequality does. If we want the strict inequality, we can modify the argument that I gave by choosing $m>n$ so that $1/m<\epsilon/2$ and $d(x_m,x)<\epsilon/2$ and then replacing $(1)$ by $$d(x_n,x)\le d(x_n,x_m)+d(x_m,x)<\frac1n+\frac1m+\frac{\epsilon}{2}<\frac1n+\epsilon.$$ Since $\epsilon$ can be chosen arbitrarily small, we conclude that $d(x_n,x)\le 1/n$ for each $n$.
A: I think it worth settling some terminology. This is from Topology: a first course by James R. Munkres. He identifies three versions of compactness for a topological space: (A) compactness, then (B) your property (which he calls "limit point compactness") and a milder condition (C) he calls "sequential compactness" which is that every infinite sequence of points has a convergent subsequence. He proves that (A) implies (B) implies (C). He also proves that the three conditions are equivalent for a metric space. 
Finally, a well-known theorem is that a metric space is compact if and only if it is complete and totally bounded. So the strongest condition is what is usually discussed in this setting. However, (A) then implies (B).
Finally, and this is not obvious, the product of two compact topological spaces is compact, and the product of two sequentially compact spaces is sequentially compact. What is unexpected is that the product of two limit point compact spaces need not be limit point compact. However, if they are metric spaces, the result does hold then.
A: To address the first question (the other is addressed in the post by Brian).
[Also, there's no error.]
Consider $\epsilon=1/(2n)$. Then there is some $N>0$ such that $d(x_m,x)<1/(2n)$ for all $m \geq N$. Let $m=\max\{N,2n\}$. Then $d(x_n,x) \leq d(x_n,x_m)+d(x_m,x) < [(1/n)+(1/m)]+1/(2n) \leq (1/n) + (1/2n) + (1/2n) = 2/n$.
