An element of a set with a finite cover must be an element of at most two open intervals in a subcover? Prove:
If a set $A\subseteq\mathbb{R}$ has a cover consisting of a finite number of open intervals, then A has a subcover such that for each $x\in A$, x is an element of at most two of the open intervals in the subcover.
My attempt:
To be honest, I have grappled with this problem for too long; I have no idea how to approach this proof. I only have the definitions of cover, subcover, and compact sets and the Heine-Borel Theorem at my disposal. I am having difficulty connecting this ideas to prove what needs to be proven. Could someone give me an idea on how to begin this proof?

What if you argued by contradiction? This is a super crude discussion on how I'm thinking one could proceed:
Suppose the statement is false. So assume that for every subcover $T'$, there exists an element $x\in A$ such that $x$ is in at least $3$ of the open intervals of the arbitrary subcover $T'$. Without loss of generality, suppose that there are exactly $3$ intervals containing $x$. 
Then, proceed to argue as @bof suggests: refine these three intervals such that one is covered by the other two and discard it from $T'$. Then, note that what remains must be another subcover of $A$ where every element $x\in A$ is in at most two intervals. This contradicts the assumption that every subcover of $A$ contains an $x$ from A that is in more than two open intervals in the subcover. 
