Baby Rudin theorem 2.41

I am reading Baby Rudin chapter 2 and came across some question on Theorem 2.41. When Rudin tries to prove that every infinite subset of $E$ has a limit points in $E$ implies $E$ is closed, he first supposes $E$ is not closed. Then there exists $x_0$ in $\mathbb{R}^k$ which is a limit point of $E$ but not a point in $E$. Then he constructs the set $S$. But later, he introduces $y$ and applies triangular inequality. I do not understand what $y$ is, and also, how he gets the last inequality of greater than or equal to $\frac12|x_0-y|$?

For reference, here is the proof given by Rudin:

2.41 $\ \ $ Theorem $\ \ $ If a set $E$ in ${\bf R}^k$ has one of the following three properties, then it has the other two:

$\quad(a)\ \ $ $E$ is closed and bounded.
$\quad(b)\ \ $ $E$ is compact.
$\quad(c)\ \ $ Every infinite subset of $E$ has a limit point in $E$.

Proof $\ \ $ If $(a)$ holds, then $E\subset I$ for some $k$-cell $I$, and $(b)$ follows from Theorems $2.40$ and $2.35$. Theorem $2.37$ shows that $(b)$ implies $(c)$. It remains to be shown that $(c)$ implies $(a)$.
$\qquad$ If $E$ is not bounded, then $E$ contains points ${\bf x}_n$ with $$|{\bf x}_n|>n\qquad(n=1,2,3,...).$$ The set $S$ consisting of three points ${\bf x}_n$ is infinite and clearly has no limit point in ${\bf R}^k$, hence has none in $E$. Thus $(c)$ implies that $E$ is bounded.
$\qquad$ If $E$ is not closed, then there is a point ${\bf x}_0\in {\bf R}^k$ which is a limit point of $E$ but not a point of $E$. For $n=1,2,3,...,$ there are points ${\bf x}_n\in E$ such that $|{\bf x}_n-{\bf x}_0|<1/n$. Let $S$ be the set of these points ${\bf x}_n$ Then $S$ is infinite (otherwise $|{\bf x}_n-{\bf x}_0|$ would have a constant positive value, for infinitely many $n$), $S$ has ${\bf x}_0$ as a limit point, and $S$ has no other limit point in ${\bf R}^k$. For if $\mathbf y\in{\bf R}^k$, ${\bf y}\neq{\bf x}_0$, then $$\begin{align}|{\bf x}_n-{\bf y}| & \geq |{\bf x}_0-{\bf y}| - |{\bf x}_n-{\bf x}_0| \\ & \geq |{\bf x}_0-{\bf y}|-\dfrac1n\geq \dfrac12|{\bf x}_0-{\bf y}|\end{align}$$ for all but finitely many $n$; this shows that $\bf y$ is not a limit point of $S$ (Theorem $2.20$).
$\qquad$ Thus $S$ has no limit point in $E$; hence $E$ must be closed if $(c)$ holds.

$\qquad$ We should remark, at this point, that $(b)$ and $(c)$ are equivalent in any metric space (Exercise $26$) but that $(a)$ does not, in general, imply $(b)$ and $(c)$. Examples are furnished by Exercise $16$ and by the space ${\scr L}^2$, which is discussed in Chap. 11.

Solution 1:

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!