Converse of Heine-Borel I am having some troubles understanding one bit about the converse of Heine-Borel which is the following statement:

Let $S$ be a subset of $\mathbb R^n$. Then the following three statements are
  equivalent:
a) $S$ is compact.
b) $S$ is closed and bounded.
c) Every infinite subset of $S$ has an accumulation point in $S$.

Here is the proof presented from the book that I am reading from I have problems in particular the implication $c) \Rightarrow b)$. I don't understand why that it is sufficient to prove that $x$ is the only accumulation point? But then according to that if I am understanding it correctly then every closed bounded set should have only 1 accumulation point which isn't true maybe I am confused here is picture of the proof. If someone could clarify the issues at which I am confused in that would be perfect.

 A: Note that $x$ isn't the only accumulation point of the (very general) set $S$, it is the unique accumulation point of the Cauchy sequence $T$. And a Cauchy sequence can only have one accumulation point.
They want to show that $S$ is closed, and they do that by showing that $S$ contains any of its accumulation points. This means taking an arbitrary accumulation point $x$ of $S$ and show that it is contained in $S$. They do this by constructing $T$, an infinite subset of $S$ with $x$ as its only accumulation point (as a subset of $\Bbb R^n$).
Since $T$ is an infinite subset of $S$, there must be some accumulation point of $T$ contained in $S$, and there is only one possibility. Therefore $x\in S$, and since $x$ was an arbitrary accumulation point, we're done.
A: The proof follows these steps:


*

*Pick an accumulation point of $S$ and call it $x$.

*Take an injective sequence $S\ni x_n\to x$, being careful never to pick $x$, and call $T^{(x_n)_1^\infty}:=\{t\in S:\exists n\in\mathbb N\ x_n=t\}$

*Use the fact that $\#T^{(x_n)_1^\infty}=\aleph_0$ and compacity of $S$ to pick $y\in S$ an accumulation point for $T$

*Use the fact that $T^{(x_n)_1^\infty}$ is the image of a sequence of points converging to $x$ to show that $y=x$. So $x\in S$.
As you can see, $T$ obviously depends on the sequence $(x_n)_1^\infty$, which depends on $x$. So there's no ground to hypothesize that this proves that $x$ is unique. Another accumulation point $x'$ simply would yield a different $T'$.
