Q: Heine-Borel Theorem From Rudin's PMA 3rd In Theorem 2.41 on page 40, to show that a compact set is bounded, it is assumed that it is not. Since it is not bounded, it must contain points $\mathbf{x}_n$ with
$|\mathbf{x}_n|>n, \,\,\,n=1,2,\dots$
According to the text, the set $S$ consisting of these points $\mathbf{x}_n$ is infinite and clearly has no limit point in $\mathbb{R}^k$.
Can someone please explain to me how $S$ "clearly" has no limit point in $\mathbb{R}^k$? To give a specific example of what is confusing me, say $n=1$, then every neighborhood of $\mathbf{x}_n = (1,1,\dots,1)$ has a point $\mathbf{q}\neq \mathbf{x}_n$ such that $|\mathbf{q}|>1 \Rightarrow \mathbf{q}\in S$.So how is that $\mathbf{x}_n$, which is in $\mathbb{R}^k$, is not a limit point?
 A: For a set A to be bounded, it is essential that all the points of A can be placed in a ball". But here you have a countable collection outside any open ball you can conjure. Thus it contradicts.
This is the logical flow: You want to show: Every inf subset has a limit point implies bounded (closed is afterwards). 
So lets proceed via Reductio Ad absurdum. You have every inf subset has a limit point. Assume your set, say A, is unbounded.  Then you can pick points $x_n \in A$ such that $|x_n| > n$ (If the set were bounded, you would stop for some n). Hence you get a countable sequence $x_n$. But this is an infinite subset of A. It must have a limit point, but it doesn't as this sequence diverges. Contradiction.
Hope that clarifies it. A key thing in the proof is that you can pick your $x_n$ distinct. If you could not, you have a finite set that is trivially bounded.
A: If $\{\mathbf{x}_n\}$ has limit point, then there exists $\mathbf{p}$ such that $B(\mathbf{p} , 1)$ contains elements of $\{\mathbf{x}_n\}$ infinitely many. And there exists natural number $N$ satisfy that $|\mathbf{p}|<N$.But if $n>N+2$, then $\mathbf{x}_n $ does not cotained $B(\mathbf{p} , 1)$. So $B(\mathbf{p} , 1)$ must contain elements of $\{\mathbf{x}_n\}$ finitely many, it is contradicted that $B(\mathbf{p} , 1)$ contains elements of $\{\mathbf{x}_n\}$ infinitely many. 
A: If $x$ is a limit point of $S$, we can find a subsequence of $\{x_{n}\}$ converging to $x$. Then that subsequence is bounded. But no subsequence of $\{x_{n}\}$ is bounded.
