Question: I am trying to prove that every bounded, infinite set has at least one limit point through the concept of open covers and without using Bolzano Weierstrass. I have no idea where to start. So far I've only assumed that a set A is bounded and infinite, so for some real number M, the set A is entirely contained in the closed and bounded set [-M, M]. By Heine Bornel I know that each open cover for [-M, M] has a finite subcover, but honestly I don't know where to go from there. Am I in the right direction?

Answer: 
Suppose $S$ is an infinite set contained in the interval $I=[-M,M]$. If $S$ has no limit points in interval $I$, then for every $x\in [-M,M]$ there is $r\_x>0$ such that the ball $B(x,r\_x)$ contains only finitely many points of $S$.

Notice that the family $\{B(x,r\_x):x\in [-M,M]\}$ is an open cover of $[-M,M]$, so there are points $x\_1,\ldots,x\_n$ such that

$$[-M,M]\subseteq \bigcup\_{i=1}^n B(x\_i,r\_{x\_i}).$$

Now, by piggeonhole principle, $S\cap B(x\_i,r\_{x\_i})$ must be infinite for some $i=1,\ldots,n$. This contradicts the choice of $r\_{x\_i}$.
You can suppose the interval is $[0,1]$. If $A$ has no limit point, then for each $x\in[0,1]$ you can find its open neighborhood $U\_x$ such that $U\_x\cap A = \{x\}$. These from an open cover, so from compactness there are $U\_{x\_1}, \ldots, U\_{x\_n}$ covering $[0,1]$. This means $A = \{x\_0, \ldots, x\_n\}$, contradiction.