In the proof, Rudin shows that $S$ has a limit point $x_0$. However, at the start of the proof we find that $x_0 \in \mathbb{R}^k - E$ and as $S \subset E$ we have found an infinite subset of $E$ which doesn't have a limit point in $E$. Well, not quite yet. It might be the case that $S$ has another limit point $y$ which is not equal to $x_0$ and is in $E$. So we must first exclude this possibility to be able to conclude that we have found an infinite subset $S \subset E$ which has no limit point in $E$. The string of inequalities that follow in Rudin's proof is to show that $x_n$ can't be simultaneously getting close to $x_0$ and $y$. I suppose the hard part is the last part:
$$\vert x_0 - y \vert - \frac{1}{n} \geq \frac{1}{2} \vert x_0 - y \vert.$$
It is important to read what follows: "for all but finitely many $n$". If $n$ is large enough (and therefore, $\frac{1}{n}$ is very small) we would essentially have
$$\vert x_0 - y \vert \geq \frac{1}{2} \vert x_0 - y \vert.$$
Which should be very clear to why it is true. In fact, we find that for $n \geq \frac{2}{\vert x_0 - y \vert}$ (well defined as $x_0 \neq y$) the first inequality holds and for $n < \frac{2}{\vert x_0 - y \vert}$ (which are finitely many) it doesn't hold. So what exactly have we shown about $y$? We have shown that $\vert x_n - y \vert$ isn't getting smaller and smaller as it should be if $y$ were a limit point of $S$. So $S$ doesn't have any other limit point besides $x_0$ and as $x_0 \notin E$ we have constructed an infinite set $S \subset E$ which has no limit point in $E$. However, this is a contradiction as we assumed c) to be true to prove a). Therefore, $E$ must be closed.
Hope this helps!
We need to show that (c) implies (a).
Suppose that every infinite subset of $E$ has a limit point in $E$. We show that then $E$ is both closed and bounded.
Suppose, if possible, that the set $E$ is unbounded. Then there exists a point $x_1 \in E$ such that $\vert x_1 \vert >1$, for otherwise the set $E$ would be contained in the unit closed ball about the origin in $\mathbb{R}^k$.
Using the same reasoning, we can find a point $x_2 \in E$ such that $$\vert x_2 \vert > 1 + \max \left( 2, \vert x_1 \vert \right).$$
Having chosen the point $x_{n-1}$ (where $n \geq 3$), we can choose a point $x_n \in E$ such that $$\vert x_n \vert > 1 + \max \left( n , \vert x_1 \vert, \ldots, \vert x_{n-1} \vert \right).$$ Otherwise, the set $E$ would be contained in a closed ball of radius equal to $1 + \max \left(n, \vert x_1 \vert, \ldots, \vert x_{n-1} \vert \right)$ and centered at the origin.
We have thus inductively chosen a sequence $\{x_n \}_{n \in \mathbb{N}}$ of distinct points of $E$ such that, for each $n \in \mathbb{N}$, we have $\vert x_n \vert > n$ and $\vert x_n \vert > \vert x_i \vert$ for all $i \in \{\ 1, \ldots, n-1 \ \}$.
Let us define the set $S$ as $$S \colon= \{ \ x_n \ \colon \ n \in \mathbb{N} \ \}.$$
This set $S$ is an infinite subset of $E$. We show that this set $S$ has no limit points in $\mathbb{R}^k$ and hence no limit points in $E$.
For any $m, n \in \mathbb{N}$ such that $n > m$, we have $$\vert x_n - x_m \vert \geq \vert x_n \vert - \vert x_m \vert \geq 1 + \max \left( n, \vert x_1 \vert, \ldots, \vert x_{n-1} \vert \right) - \vert x_m \vert > 1.$$
Thus it follows that, for any $m, n \in \mathbb{N}$ such that $n \neq m$, the inequality $\vert x_n - x_m \vert > 1$ holds.
So if some point $x \in \mathbb{R}^k$ were a limit point of $S$, then there would be infinitely many values $n \in \mathbb{N}$ such that $$\vert x_n - x \vert < \frac 1 4,$$ and for any two (distinct) such points $x_m$ and $x_n$ of $S$, we would have $$\vert x_m - x_n \vert \leq \vert x_m - x \vert + \vert x_n - x \vert < \frac 1 4 + \frac 1 4 = \frac 1 2,$$
which contradicts what we have shown above about the distance between any two distinct points of $S$.
So the set $S$, though an infinite subset of $E$, fails to have a limit point in $\mathbb{R}^k$ and hence in $E$.
Therefore, the set $E$ must be bounded.
Next, suppose that $E$ is not closed. Then $E$ has a limit point $x_0 \in \mathbb{R}^k - E$. Since $x_0$ is a limit point of $E$, every neighborhood of $x_0$ contains a point of $E$ distinct from the point $x_0$ itself (in fact infinitely many points of $E$).
Thus, there is a point $x_1 \in E$ such that $$0 < \vert x_1 - x_0 \vert < \frac 1 2.$$
Again there is a point $x_2 \in E$ such that $$0 < \vert x_2 - x_0 \vert < \min \left( \vert x_1 - x_0 \vert, \frac 1 3 \right).$$
Assuming that the point $x_{n-1}$ (where $n \geq 3$) has been chosen, we can choose a point $x_n \in E$ such that $$0 < \vert x_n - x_0 \vert < \min \left( \vert x_1 - x_0 \vert, \ldots, \vert x_{n-1} - x_0 \vert, \frac{1}{n+1} \right).$$
Thus we have recursively defined a sequence $\{x_n \}_{n\in\mathbb{N}}$ of points of $E$ for which $x_n \neq x_m$ for all $m, n \in \mathbb{N}$ such that $m \neq n$ and also $$0 < \vert x_n - x_0 \vert < \frac 1 n \ \mbox{ for all } \ n \in \mathbb{N}.$$
Let us define the set $S$ as follows:
$$S \colon= \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}.$$
This set $S$ is an infinite subset of $E$.
We show that $x_0$ is the only limit point of $S$. That is, we show that $x_0$ is a limit point of $S$ but no other point $y$ of $\mathbb{R}^k$ can be a limit point of $S$.
Let $\delta$ be any positive real number. Then, by the archimedean property of $\mathbb{R}$, we can find $n_\delta \in \mathbb{N}$ such that $$n_\delta > \frac 1 \delta,$$ and so, for all $n \in \mathbb{N}$ such that $n \geq n_\delta$, we have $$0 < \vert x_n - x_0 \vert < \frac{1}{n+1} < \frac 1 n_\delta < \delta,$$
which implies that $x_0$ is indeed a limit point of $S$.
Now if $y \in \mathbb{R}^k$ and $y \neq x_0$, then $\vert y - x_0 \vert > 0$. So we can find a positive integer $N$ such that $$N > \frac{2}{\vert y -x_0 \vert}.$$
So, for every $n\in \mathbb{N}$ such that $n \geq N$, we have
$$ 0 < \vert x_n - x_0 \vert < \frac{1}{n} \leq \frac 1 N < \frac{\vert y - x_0 \vert}{2}$$
and hence, for every $n\in \mathbb{N}$ such that $n \geq N$, we have
$$\vert x_n - y \vert \geq \vert y - x_0 \vert - \vert x_n - x_0 \vert \geq \vert y - x_0 \vert - \frac{\vert y - x_0 \vert}{2} > \frac{\vert y - x_0 \vert}{3}. $$
So if we take a positive real number $\epsilon$ such that
$$0 < \epsilon < \frac{1}{2} \min \left( \vert x_1 - y \vert, \ldots, \vert x_N - y \vert, \frac{\vert y - x_0 \vert}{3} \right),$$
then there is no point of set $S$ that lies in the neighborhood of the point $y$ of radius $\epsilon$, other than the point $y$ itself if $y \in S$; that is,
$$S \cap \left( N_\epsilon (y) - \{ y \} \right) = \emptyset,$$
which implies that the point $y$ cannot be a limit point of the set $S$.
But $y$ was any point of $\mathbb{R}^k$ other than the point $x_0$. Therefore, $x_0$ is the only limit point of the set $S$.
But $x_0 \not\in E$ by our hypothesis. Thus we have found an infinite subset $S$ of $E$ such that no point of $E$ is a limit point of $S$. The only limit point of $S$, namely the $x_0$, ( which is also a limit point of the set $E$ by our hypothesis) does not belong to $E$.
So if every infinite subset of the set $E \subset \mathbb{R}^k$ were to have a limit point in $E$, then the set $E$ must be closed and bounded.
Best Answer
To elaborate on the "otherwise", let us look at the negation.
That is, let us ask ourselves what would happen if that were not true.
Suppose that $S$ were finite.
Note that $S$ was constructed in a way such that we have the following property:
Now, since $S$ is finite, this means that not all $x_n$ can be distinct. In fact, it would mean that there's some $x' \in S$ such that $$x' = x_n\quad\text{for infinitely many } n \in \Bbb N.$$
(Why? If that didn't happen, then $\Bbb N$ could be written as a finite union of finite sets.)
Let $A$ be the collection of all such $n$ that satisfy the above condition. This is an infinite set. What Rudin is saying is that
$$|x_n - x_0| \text{ is consant for all } n \in A.$$ And that is simply because $|x_n - x_0| = |x' - x_0|$ for all $n \in A$.
This is why $|x_n - x_0|$ is constant for infinitely many $n$.
Moreover, this constant is positive because $x_0 \notin S \ni x'$ and thus, $x' \neq x_0$.
The contradiction that he would derive from this is that a fixed positive number $\delta = |x' - x_0|$ is less than $1/n$ for infinitely many $n$. (Since it's infinitely many, you choose an $n > 1/\delta$ and arrive at the contradiction.)