Proving that a set in which any sequence having exactly one accumulation point converges is compact I found the following question about compactness:

A set $S\subseteq \mathbb{R}^n$ has the following property:
  if a sequence of $S$ has exactly one accumulation point in $S$, then it converges in $S$. Prove that $S$ is compact. 

We know that in $\mathbb{R}^n$ compact sets correspond to closed and bounded sets. 
(1) I am trying to prove that $S$ is closed. The condition in the question says if a sequence of $S$ has an accumulation point in $S$ then it converges in $S$. I think instead the condition should be if a sequence of $S$ has an accumulation point( in $\mathbb{R}^n$, not necessarily in $S$ )then it converges in $S$.
Am I correct?
(2) How to prove that $S$ is bounded?
P.S. I have made an edit (there is exactly one accumulation point).
Source of the question: Question 1 of this problem set.  
 A: With the edit, the question makes more sense now. 
Now rather than proving $S$ is closed and bounded, you could try proving an equivalent characterization: any sequence has a converging subsequence (it is equivalent in a finite dimensional normed vector space, and equivalent to the "real" definition of compactness in any metric space). I think it would be easier.
I will assume $S\neq \emptyset$ (the case $S=\emptyset$ being trivial).
Assume $(s_n)$ is a sequence in $S$ with no converging subsequence in $S$. In particular it has no accumulation point. 
Let $x\in S$ be any point and let $x_{2n} = x$, $x_{2n+1} = s_n$. Since $(s_n)$ has no accumulation point, it follows that $x$ is the only accumulation point of $(x_n)$. 
According to the hypothesis, $(x_n)$ converges, which in particular implies that $(s_n)$ converges (as it is a subsequence), which contradicts the assumption that $(s_n)$ had no converging subsequence. 
Therefore, $(s_n)$ has a converging subsequence. Such a subsequence has exactly one accumulation point, and it takes values in $S$, therefore, again according to the hypothesis, it converges in $S$. 
$(s_n)$ being arbitrary, we proved: any sequence with values in $S$ has a converging subsequence in $S$.
Thus, $S$ is compact.
A: Using the fact that for a metric space, compactness and sequentially compactness are equivalent.
Let $(a_{n})$ be an arbitrary sequence of real numbers in $S$. Then clearly by the hypothesis $(a_{n})$ has an accumulation point in $S$ say $a$. 
Then we will construct a convergent subsequence of $(a_{n})$ converging to $a$.
$\because a$ is an accumulation point every neighborhood of $a$ has infinitely many points of $(a_{n})$.
$\therefore B(a,1)$ contains $a_{n_{1}}$.
Further, $B(a,1/2)$ contains infinitely elements of $(a_{n})$. Therefore $\exists \ a_{n_{2}}\in B(a,1/2)$ such that $n_{1}<n_{2}$.
Similarly, $B(a,1/k)$ contains infinitely elements of $(a_{n})$. Therefore $\exists \ a_{n_{k}}\in B(a,1/k)$ such that $n_{k-1}<n_{k}$
Therefore,$\exists$ a subsequence $(a_{n_{k}})$ of $(a_{n})$ converging to $a$.
Hence, every sequence of set $S$ has a convergent subsequence. 
