If $K$ is compact then it is bounded. (that it is closed was very easy to prove)
Definitions
.A set is called compact if every sequence has a convergent subsequence.
Attempt
Let $K$ be compact and assume that it is not bounded. Let $x \in K$ be any point. Then for every $n \in \mathbb N$ let $x_n$ be such that $d(x,x_n) > n$.
Now I want to show that $x_n$ does not have a convergent subsequence. I tried by contradiction: assume $x_{n_k}$ was a convergent subsequence and $x_{n_k}\to y$ for some $y \in K$. How to proceed?
Best Answer
Alternatively you can use the equivalently definition that :
$K$ is compact $\iff $ Every open cover of K has a finite subcover
Let $U_1(x)$ be a ball with radius $1$ around x. Then cover your set with those balls and use the compactness to get a finite cover of those balls with radius 1. Its easy to conclude now that your set is bounded above.
Edit: (Second Proof)
Let's assume that K is unbounded. We now want to construct a sequence which does not have a convergent subsequence in K.
Let $x_0\in K$ be arbitrary. We choose $x_1\in K$ such that $d(x_1,x_0)\geq1$. We will define the next parts of the sequence recursively:
If we already got $x_0,x_1,....,x_k$ then $R_k$ is the radius of a circle about $x_0$ which contains $x_1,....,x_k$. We choose now $x_{k+1}$ such that $d(x_{k+1},x_0)\geq R_k+1$. For all $l \leq k$ we conclude:
$d(x_{k+1},x_l)\geq d(x_{k+1},x_0)-d(x_0,x_l)\geq R_k+1-R_k=1 $
So, any two parts of $(x_n)_{n\in \mathbb N}$ have a distance at least 1 to each other, hence this sequence does not contain a convergent subsequence.