Questions about quantifiers and justification for certain assumption in a proof about limit points Let $E$ be an unbounded subset of $\mathbb R^k.$ Then we can construct an infinite  $S \subset E$ with no limit point in $E.$ The process is given below:
 

I have questions about the outlined parts above.
Orange outline:
Shouldn't it be "$\exists n \in \mathbb N, \exists x_n \in E$ s.t. $|x_n| < n$" instead of what's there?
Blue outline:
Is that a way for us to talk about the neighborhoods of $p$? There's no (?) neighborhood of $p$ where $N < |p|$ if we take $N$ to be a radius around $p$ so that we only consider $N \ge |p|.$ The reason I ask this is because it looks like some random assumption thrown in to get $|x_n - p| \ge 1,$ but it must've come from some considerations. I am trying to logically tie it to other assumptions that precede it.
 A: The first paragraph lays out the construction of $S$. Informally speaking, $S$ turns out to be sort of like the opposite of a perfect set in the context of topology (I have some facetious terminology ideas that I shall spare you from).
The second paragraph shows via contradiction that $S$ is an infinite set. The argument being that if $S$ were a finite set then $ m= \displaystyle \max_{\mathbf{x} \in S} |\mathbf{x}|$ would exist ($m$ is not necessarily an integer). And so $n=m+1$ would have to be an upper bound for the set $A=\{|\mathbf{x}|: \mathbf{x} \in E\}$ which symbolically translates to
\begin{align}\exists n \in \mathbb{R}_+:\forall \mathbf{x} \in E \,\big(|\mathbf{x}| \leq n \big) .
\end{align} 
This satisfies the definition for a bounded subset of $\mathbb{R}^k$, and so we have contradicted the fact that $E \subset \mathbb{R}^k$ is not bounded. Therefore $S$ is an infinite set.
After the first two paragraphs, the author shows that there does not exist $\mathbf{p} \in \mathbb{R}^k$ such that $\mathbf{p}$ is a limit point of $S$ ($E \subset \mathbb{R}^k$). The negation of the sentence
\begin{align}\exists \mathbf{p} \in \mathbb{R}^k:\mathbf{p} \text{ is a limit point of } S
\end{align} is the sentence
\begin{align}\forall \mathbf{p} \in \mathbb{R}^k:\mathbf{p} \text{ is not a limit point of } S .
\end{align} 
To this end, the author fixes an arbitrary point $\mathbf{p} \in \mathbb{R}^k$. Since $\{\mathbf{p}\} \subset \mathbb{R}^k$ is a singleton, we know rather trivially that it is a bounded subset of $\mathbb{R}^k$. Therefore we may let $N$ denote the smallest positive integer greater than $|\mathbf{p}|$ (well-ordering principle).
A: Here is a generalization to all metric spaces.
If A is an unbounded subset of a metric space (S,d), then
there is an infinite subset of A without limit points.
Proof.
Since A is unbounded, it's not empty;  so pick any p1 from A.
Proceed by induction, having picked p1,.. p_n from A, pick
some p_(n+1) from A - B(p1, d(p1,p_n) + 1) which is possible
since A is unbounded.  
Let d_n = d(p1,p_n).  Thus 0 = d_1 and for all n, d_n + 1 < d_(n+1).
Whence {d_n} is a strictly increasing unbounded (above) sequence.  
Let K = { p_n | n in N }.  K is infinite because were it finite,
the sequence {d_n} would have an upper bound, which it doesn't.  
Let a be any point.
To show a is not a limit point of K, let da = d(p1,a).
As {d_n} is unbounded, there's some integer n with da < d_n.
Let k be the least integer with da < d_k.  Thus d_(k-1) <= da < d_k.
Let r = d_k - da;  0 < r.  Let s = da - d_(k-1);  0 < s.
For all n >= k, d_k <= d_n <= da + d(a,p_n);  r <= d(a,p_n).
For all n < k - 1 (if any), da <= d_n + d(p_n,a) < d_(k-1) + d(p_n,a);
. . s < d(p_n,a).  
Thus B(a, min {r,s}) misses K except possibly at p_(k-1).
If a /= p_(k-1), then B(a, min{ r,s, d(a,p_(k-1) }) misses K.
If a = p_(k-1), then B(a, min{ r,s }) has no other point of K.
Either way, a is not a limit point of K.  
