Prove $C$ is compact $\iff$ every sequence in $C$ has a subsequence that converges to a point in $C$ I am able to prove the direction from Compact to subsequence but I can't seem to get all the other direction quite right.
I can prove that $C$ must be bounded, but then my professor says that I am not correct about why it must be closed.  I said if $C$ is open then there must exist a convergent sequence in $C$ such that the limit is not in $C$ which is a contradiction to the assumption which would imply that $C$ is closed.
Can someone explain what I am doing wrong?
 A: It seems that $C \subseteq \mathbb{R}^n$ and you wish to prove that $C$ is bounded and closed, assuming that every sequence in $C$ has a convergent subsequence with the limit in $C$.
If $C$ is not bounded, then there exists a sequence $(x_n)_n$ in $C$ such that $\|x_n\| \ge n, \forall n \in \mathbb{N}$. However, every subsequence of $(x_n)_n$ is unbounded, and hence cannot converge to an element of $C$.
To show that $C$ is closed, assume $(x_n)_n$ is a sequence such that $x_n \xrightarrow{n\to\infty} x$, where $x \in \mathbb{R}^n$. We wish to prove $x\in C$.
There exists a subsequence $(x_{p(n)})_n$ such that $x_{p(n)} \xrightarrow{n\to\infty} y \in C$. Since every subsequence of a convergent sequence also converges to the same limit, we conclude $x = y \in C$.
Hence, $C$ is closed.
A: If $C$ is not compact, then it is not both closed and bounded.
If $C$ is not bounded, then, for each natural $n$, take $x_n\in C$ such that $|x_n|>n$. Then the sequence of al  $x_n$'s has no convergent subsequence.
If $C$ is not closed, let $x\in\overline C\setminus C$. Take a sequence $(x_n)_{n\in\mathbb N}$ of elements of $C$ which converges to $x$. Then the sequence $(x_n)_{n\in\mathbb N}$ has no convergent subsequence. 
