Theorem: Show that an infinite set $\displaystyle S$ is compact if and only if every infinite subset of $\displaystyle S$ has an accumulation point that lies in $\displaystyle S$.

My attempt of proof:

Suppose $\displaystyle S$ is infinite and compact but that there exists an infinite subset that does not contain any accumulation points in $\displaystyle S$. Call this subset $\displaystyle S'$ and let an accumulation point of this subset be $\displaystyle a$. Since $\displaystyle a$ is an accumulation point of $\displaystyle S'$, there is a sequence of points in $\displaystyle S'$ such that the limit of that sequence is $\displaystyle a$. Clearly this sequence lies in $\displaystyle S$ as well since $\displaystyle S' \subset S$, so we have that $\displaystyle a$ is an accumulation point of $\displaystyle S$, and since $\displaystyle S$ is compact, $\displaystyle a\in S$, which is a contradiction. So all infinite subsets of $\displaystyle S$ have an accumulation point which lies in $\displaystyle S$.

Now suppose every infinite subset of $\displaystyle S$ has an an accumulation point in $\displaystyle S$ and that $\displaystyle S$ is not compact. Then $\displaystyle S$ is either unbounded or not closed. If $\displaystyle S$ is unbounded, then there exists a sequence in $\displaystyle S$, say $\displaystyle \{x_{n}\}$, such that for all $\displaystyle n\in \mathbb{N}$, $\displaystyle x_{n} \ge n$, so $\displaystyle \lim x_{n} = \infty$. So let $\displaystyle S''$ be the set of these sequence elements. By construction, this is an infinite set, yet it has no accumulation point since no matter the ordering of the terms, there is no such point $\displaystyle b$ such that infinitely many sequence terms lie arbitrary close to $\displaystyle b$. This contradictions our hypothesis, so $\displaystyle S$ is bounded. Suppose $\displaystyle S$ is not closed. Then there exists a point $\displaystyle c$ such that $\displaystyle c\in \partial S$ yet $\displaystyle c\not\in S$. Since $\displaystyle c\in \partial S$, there exists a sequence of points in $\displaystyle S$, say $\displaystyle \{y_{n}\}$ such that $\displaystyle \lim y_{n} = c$. Let $\displaystyle S'''$ be the set of these sequence elements. By construction, this is an infinite set, yet its only accumulation point is $\displaystyle c$ since no matter the ordering of the terms, $\displaystyle c$ is the only point where infinitely many terms of the sequence are arbitrarily close to. Again, this contradicts the hypothesis, so $\displaystyle S$ must be compact. $\displaystyle \blacksquare$

Is this proof valid, the only part of the proof I am really unsure of is in the second paragraph with the construction of those sequences. Is my claim valid that no matter the ordering, there are no other accumulation points then the ones I have specified? It seems obvious/true to me (I think) but is it a false claim or is it a claim that needs proof? Thank you.