Something about the proof of Bolzano-Weierstrass Theorem not making sense So I'm trying to understand the proof of this version of the theorem

Here is the proof given

I have understood Lemma 1. and its implications.
The thing I don't get about it is:
We assume $S$ has no limit point, but then out of nowhere, because $\inf(X)\in X$ then $S$ has a limit point. Doesn't this contradict our assumption straight away?
My specific questions are:
1)Why does $X\subset S \Rightarrow \inf(X)\in X$ (and $S$ for the matter).
2)Why do we let $\inf(X)$ be a limit point? Wouldn't that go against the whole point of the proof?
 A: For any subset $A$ of $\Bbb R$ with a lower bound, if $\inf A\notin A$, then $\inf(A)$ is a limit point of $A$. That's so because, for any $n\in\Bbb N$, there is some $a_n\in\left(\inf(A),\frac1n+\inf(A)\right)\cap A$, by the definition of $\inf(A)$ and because $\inf A\notin A$. But then $\lim_{n\to\infty}a_n=\inf(A)$ and therefore $\inf(A)$ is a limit point of $A$.
So, if $\inf(X)\notin X$, then $\inf(X)$ is a limit point of $X$ and therefore a limit point of $S$. But we are assuming that $S$ has no limit points. So, $\inf(X)\in X$.
A: I guess that Lemma 1 says
If $S$ is a bounded set of real numbers such that, for every nonempty subset $X$ of $S$, $\inf(X)\in X$ and $\sup(X)\in X$, then $S$ is finite.
Now your task is to show that if $S$ is a bounded set with no limit point, then $S$ is finite. Using Lemma 1 requires to consider nonempty subsets of $S$, so pick one and call it $X$.
Then we know that $\inf(X)$ exists. If $\inf(X)\notin X$, then it is a limit point of $X$. Indeed, for every $\varepsilon>0$, there exists $x\in X$ such that
$$
x<\inf(X)+\varepsilon
$$
and therefore $(\inf(X)-\varepsilon,\inf(X)+\varepsilon)$ intersects $X$ in (at least) a point different from $\inf(X)$.
Similarly, if $\sup(X)\notin X$, then $\sup(X)$ is a limit point of $X$.
On the other hand, a limit point of $X$ is also a limit point of $S$, which has none by assumption. Therefore $\inf(X)\in X$ and $\sup(X)\in X$.
Since $X$ was an arbitrary nonempty subset of $S$, Lemma 1 applies and so $S$ is finite.
A: Regarding your comment "out of nowhere", what's happening here is a contradiction proof within a contradiction proof. As you say, we have assumed that $S$ has no limit point.
Next comes the following line of deduction regarding any nonempty subset $X$ of $S$: If we also assume that $\inf(X) \not\in X$, then $\inf(X)$ is a limit point of $X$ and therefore $\inf(X)$ is a limit point of $S$, and this is a contradiction. Therefore we conclude that $\inf(X) \in X$.
So we have proved: every nonempty subset of $S$ contains its own infimum.
And then, similarly, every nonempty subset of $S$ contains its own supremum.
