Every Cover of a Compact Real Interval by Open Intervals Has a Finite Subcover where only Consecutive Sets Overlap?

In fact, an open cover of $X$ where any point of $x$ is in at most $n$ elements of the cover is called a cover of order $n$, and such covers are used in the covering dimension in general topology: a space has $\dim(X) \le n$ iff every finite cover of $X$ has a refinement of order $\le n+1$. (see Wikipedia for more info.)

A theorem by Lebesgue shows that $\dim([0,1])=1$ and so refinements of order $2$ (which is what you want) exist. If we are in an ordered space and we take covers of open intervals we get our minimally overlapping sets as you desire.

This can be proved via a kind of "topological induction" argument - just like Heine-Borel itself. (See this survey paper for more on the method; there it's called "real induction.")

Consider a covering $C$ of a compact real interval $[a,b]$. Let $F$ be the set of points in $[a,b]$ which are "nicely reachable" by $C$: that is, $x\in F$ iff there is a $D\subseteq C$ which covers $[a,x]$ and has the intersection property you want relative to $[a,x]$. Clearly $a\in F$, so $F\not=\emptyset$. Now let $c=\sup(F)$. Some element of $C$ covers $c$, and thinking about how far "to the left" this element reaches we see that $c\in F$ as well. But then if $c\not=b$, that same element of $C$ lets us extend $F$ a bit "to the right" past $c$, contradicting the definition of $c$. So $c=b$ and $c\in F$. But this proves the desired claim.

I've been a bit slippery in the above argument; it's a good exercise to fill in all the steps, and in particular to formally express "the intersection property you want relative to $[a,x]$."