TITLE: Every Cover of a Compact Real Interval by Open Intervals Has a Finite Subcover where only Consecutive Sets Overlap?
QUESTION [6 upvotes]: Intuitively this seems true and would be a useful lemma in proving the fundamental theorem of calculus without assuming continuity of the derivative, i.e. that if $f$ is differentiable on $[a, b$ and the derivative is Riemann integrable then $\int_a^b f' = f(b) - f(a)$. But is it true and if so is there any standard terminology for such a cover ? My attempt at a proof follows.

Let $I = [a, b]$ be a closed bounded interval in $\mathbb R$. Then (Hiene-Borel theorem) every open cover of open intervals has a finite sub-cover. We can assume that the open intervals have distinct R-endpoints, as for any two with the same R-endpoint one must be contained within the other and the smaller one (or either if the same L-endpoint) can be discarded without affecting the cover.
Then these open intervals $\{O_i\ = (a_i, b_i)\}_{i = 1, n}$ can be ordered by their R-endpoints $\{b_i\}_{i = 1, n}$ in a strictly ascending sequence $b_1 < b_2, ...< b_n$.
Claim:   
from such a set $\{O_i = (a_i, b_i)\}_{i = 1, n}$ we can select a subset which covers $[a, b]$, renumbered as $\{O'_j = (a'_i, b'_1)\}_{j = 1, m}$ in strictly ascending sequence $b'_1 < b'_2, ...< b'_m$ such that for $|i - j| > 1$ then $O'_i \cap O'_j = \emptyset$ and for $|i - j| = 1$ then $O'_i \cap O'_j \not= \emptyset$.
Construction: 
Choose $O'_1$ from intervals $O_i$ having  $a \in O_i$ and maximizing $b_i$ among such intervals. 
Iteratively, stop if $b$ is in the last chosen interval $O'_j$, otherwise, ...
Chose $O'_{j+1}$ from intervals $O_i$ having  $b'_j \in O_i$ and maximizing $b_i$ among such intervals. 
Then the set $\{O'_j\}  $ fulfills the requirements.
Proof:
Since the set $\{O_i\}  $ covers $[a, b]$ then for $a$ in step1 and for every $b'_j$ in the iteration there is an $O_i$ which contains it and since the endpoints are unique there is exactly one which maximizes $b_j$.
Since the $O_i$ chosen as $O'_{j+1}$ contains $b'_j$ and is open then it has a non-empty intersection with $O'_j$. I.e. consecutive open intervals intersect.
The $O_i$ chosen as $O'_{j+1}$ cannot intersect any interval prior to $O'_j$ as this would require it to have been chosen previously in order to maximize $b_j$.

REPLY [3 votes]: 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 $[a,x]$."