Pretty much just what the title says.
So if a set is compact, then it is closed and bounded, correct? Since the set E is bounded, the sequence $\{x_n\}$ is also bounded. Meaning that it has a limit point (I can not remember the theorem stating that). I think that is how I should begin? But I am not sure how to proceed.
I know there was a previous post on this, but there was no complete answer. It was essentially just saying what you need to do to prove an if and only if.
A bounded sequence does not have necessarily a limit: take for example $x_n=(-1)^n$ in $\mathbb{R}$, this defines a bounded sequence but it has no limit. What is true is that it admits a converging subsequence (Bolzano-Weierstrass theorem). Using closure of $E$ you get that this limit point must be in $E$ and one implication is done. For the other implication: take $E$ with the property above about sequences. By absurd suppose it is not bounded, so for every $n \in \mathbb{N}$ there exists a point $x \in E$ such that $|x| \geq n$ and we can use this to construct easily a sequence admitting no converging subsequence (I think you can write down the details). Then in an metric space as $\mathbb{R}^n$ a subset is closed iff the limit of any converging subsequence of its elements is still in the subset. Take a sequence of elements $\{ x_n\}_n$ of $E$ converging to a point $x \in \mathbb{R}^n$, by the property of $E$ there exists a subsequence converging to a point $y \in E$. But since the original sequence already converges by uniqueness of the limit we must have $y=x \in E$ and $E$ is closed.