A subset S of $R$ is complete if and only if it is closed. I cannot for the life of me seem to understand this proof. Here is my attempt.
Proof:
"==>": If $S$ is complete, every Cauchy sequence in $S$ converges to a point $x \in S$. Then $x$ is a limit point of $S$. Thus $S$ is closed.
"<==": If S is closed, then it contains all limit points. So $x \in S$. If $x$ is a limit point, then there is a sequence $a_n$ (with $a_n$ an element of S for all natural numbers) and $a_n$ converges to $x$. So the sequence is Cauchy and converges to a point in S. Thus, S is complete. QED.
What is wrong with my proof? Thanks for the help.
Edit: my updated proof for the direction "<==" is as follows:
Let $a_n$ be Cauchy in S and so $a_n$ converges to a point $x \in R$ (by Completeness of R). (We want to show $a_n$ converges to a point in S). So x is a limit point of S. Since, by assumption, S is closed and so it contains all limit points, this implies $x \in S$. Thus, every Cauchy sequence in S converges to a point in S. So S is complete. Is this correct now?
 A: For $\implies$ you need to let $x$ be an arbitrary limit point and show it is in $S$.
Answer: Let $x$ be a limit point of $S$. Then, there is a sequence $x_n$ with each $x_i\in S\quad \forall i\in\mathbb{N}$ which converges to $x$. Since the sequence $x_n$ is convergent, it is Cauchy. Thus, since $S$ is complete by assumption, $x\in S$. Thus, since $x$ was an arbitrary limit point of $S$ we have shown that all limit points of $S$ are elements of $S$ and so $S$ is closed.
For the other direction, you need to let $x_n$ be an arbitrary Cauchy sequence and show that it converges to a point $x\in S$.
Answer: I think you can do this one on your own.
A: The main problem is that you’ve not really sorted out exactly what you need to prove. For the first part, you want to show that $S$ is closed, so you must let $x\in\Bbb R$ be an arbitrary limit point of $S$ and prove that $x\in S$.

Suppose that $S$ is complete. Let $x$ be any limit point of $S$. Then there is a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ in $S$ that converges to $x$. Every convergent sequence in $\Bbb R$, is Cauchy, and the metric in $S$ is the same as the usual metric in $\Bbb R$, so $\sigma$ is a Cauchy sequence in $S$. $S$ is complete, so $\sigma$ converges to some point $y\in S$. Finally, we know that a sequence in $\Bbb R$ cannot converge to two different points (why?), so $y=x$, and therefore $x\in S$. This shows that $S$ contains all of its limit points and therefore is closed.

For the second part you must show that every Cauchy sequence in $S$ converges to a point of $S$. Your argument clearly cannot be doing that, because you did not start with an arbitrary Cauchy sequence in $S$: you started with what you’re supposed to be finding, namely, a limit.

Suppose that $S$ is closed, and let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be any Cauchy sequence in $S$. Then $\sigma$ is a Cauchy sequence in $\Bbb R$, which we know is complete, so $\sigma$ converges to some $x\in\Bbb R$. Let $\epsilon>0$ be arbitrary. Then there is an $m_\epsilon\in\Bbb N$ such that $|x_n-x|<\epsilon$ whenever $n\ge m$. In particular, $x_{m_\epsilon}\in S\cap(x-\epsilon,x+\epsilon)$. Thus, every nbhd of $x$ contains a point of $S$, and therefore $x\in\operatorname{cl}S$. But $S$ is closed, so $\operatorname{cl}S=S$, and hence $x\in S$. This shows that every Cauchy sequence in $S$ converges to a point of $S$.

I’ve been a bit wordy in both arguments, but when you’re just starting to write proofs, it’s better to give more rather than fewer details, and I wanted to make clear exactly what the logic of the arguments is.
A: Allow me to add an answer based on the concepts from topology.
$(\impliedby)$ Let $A$ be closed in $\mathbb{R}^n$ and $(x_k)$ be a Cauchy sequence in s.t. $x_k \in A$. Since, a Cauchy sequence is bounded, $(x_k)$ is bounded and hence, by Bolzano-Weierstrass, has a convergent subsequence. A Cauchy sequence with convergent subsequence is convergent. Therefore, $(x_k)$ converges to say $x$, where $x$ is a limit point of $A$. Since, $A$ is closed, $x \in A$. Therefore, $A$ is a complete metric space.
$(\implies)$ Let $A$ be a subset of $\mathbb{R}^n$ which is a complete metric space and $x$ be a limit point of $A$. Then, $\exists$ a Cauchy sequence $(x_k)$ s.t. $x_k \in A$ and $x_k \rightarrow x$. Since, $A$ is complete, $x \in A$. Therefore, $A$ contains all of its limit points, hence is closed.
