An infinite subset of Euclidean space and its limit points My question is about the theorem 2.41 in baby Rudin, which is the following.
Theorem. $ \ $  If a set $E \subset \mathbb{R}^k$ has one of the following three properties, then it has the other two:
(a) $E$ is closed and bounded.
(b) $E$ is compact.
(c) Every infinite subset of $E$ has a limit point in $E$.
Proof $ \ $ $\cdots$. $\ $
It remains to be shown that (c) implies (a). If $E$ is not bounded, then $E$ contains points $\boldsymbol{x}_n$ with
\begin{align*}
|\boldsymbol{x}_n|>n \quad (n=1,2,3,\cdots).
\end{align*}
The set $S$ consisting of these points $\boldsymbol{x}_n$ is infinite and clearly has no limit point in $\mathbb{R}^k$, hence has none in $E$. Thus, (c) implies that $E$ is bounded.
$\cdots$
My question is, how do I know that $S$ is infinite? Also, how do I know that there is no limit point of $S$ in $\mathbb{R}^k$?
 A: Here is a (very) detailed answer to your questions. If you do some drawing, you will see that the following reasoning is natural.
By assumption, the given set $E$ is unbounded. Therefore, for any $n$, you can pick some element in $E$ outside of the ball centered at $0$ with radius $n$. Let us call this chosen point $x_n$, and let us collect all these elements into a subset $S$ of $E$. By construction, elements of $S$ form a sequence $x_1,x_2,\ldots$ with $|x_n| > n$.
Suppose by contradiction that $S$ is finite. Therefore, there exists a finite amount of points, say $y_1,\ldots,y_N$ such that $S = \{y_1,\ldots,y_N\}$. If $M = \max_{j=1\ldots,N} |y_j|$, then by definition, we have that any element of $S$ is of norm a most $M$. That is to say
$$
\forall n, \quad |x_n| \leqslant M.
$$
Since we know that $|x_n|>n$, it follows that
$$
\forall n,\quad n < M,
$$
which is impossible (take, for example, $n$ to be an integer greater than $M$). Therefore, $S$ is not finite, which means it is infinite.
Now that we know that $S$ is infinite, let us show that it cannot have any accumulation point in $\Bbb R^k$. Recall that $x \in \Bbb R^k$ is an accumulation point of $S$ if for any positive number $\varepsilon >0$, the punctured ball $B(x,\varepsilon)\setminus \{x\}$ and $S$ meet, that is if $\big(B(x,\varepsilon)\setminus \{x\}\big)\cap S \neq \varnothing$.
Let $x \in \Bbb R^k$ be any point and choose some integer $N$ such that $N > |x|+1$. Then by assumption, for all $n>N$, $|x|+1<n < |x_n|$, so that the triangle inequality gives $|x-x_n| > 1$. Therefore, only a finite amount of elements of $S$ can be very close to $x$ (at distance less than $1$): they are $x_1,\ldots,x_N$. Now, consider the least distance of these elements to $x$: $d = \min\{ d(x,x_j)~\mid~ j=1,\ldots,N, x_j \neq N\}$ (we have removed points $x_j$ that are equal to $x$). Finally, define $\varepsilon = \min\{d,1\}$. Then by construction, the punctured ball $B(x,\varepsilon)\setminus \{x\}$ does not contain any point of  $\{x_1,\ldots,x_N\}$ (since they are either equal to $x$ or at distance greater than $d$ from $x$), and does not contain any element of $\{x_n ~\mid~ n> N\}$ (since they are at distance greater than $1$ from $x$). We have finally shown that:
$$
\exists \varepsilon >0, \big(B(x,\varepsilon)\setminus\{x\}\big) \cap S = \varnothing,
$$
meaning that $x$ is not an accumulation point of $S$. This being true for any $x$, $S$ has no accumulation point in $\Bbb R^k$.
