Every Cover of a Compact Real Interval by Open Intervals Has a Finite Subcover where only Consecutive Sets Overlap? 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 $b_j$.
 A: 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.
A: 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 $[a,x]$."
A: That version of FTC is trivial:


FTC If $f:[a,b]\to\Bbb R$ is differentiable and $f'$ is Riemann integrable then $\int_a^b f'=f(b)-f(a)$.


Proof: Say $a=x_0<x_1<\dots<x_n=b$. MVT shows that there exist $\xi_j\in(x_{j-1},x_j)$ with $$f(b)-f(a)=\sum_{j=1}^n(f(x_j)-f(x_{j-1}))=\sum_{j=1}^n(x_j-x_{j-1})f'(\xi_j).$$That last sum is a Riemann sum for $\int_a^b f'$, so it is within $\epsilon$ of $\int_a^b f'$ if $\max_j(x_j-x_{j-1})<\delta$.
Heh: I sometimes conjecture that the above is why Riemann defined the integral exactly the way he did...
The thing about covers of $[0,1]$ is also not hard:


Lemma: If three open intervals have a point in common then one of the intervals is contained in the union of the other two.


The proof of  that is simple.
Now say  $I_1,\dots,I_n$ are open intervals and $[0,1]\subset\bigcup I_j$. Extract a minimal subfamily $J_1,\dots,J_m$ that still covers $[0,1]$. Minimality shows that  no point lies in more than two of the $J_k$, by the lemma; hence the $J_k$ can be ordered in the way you want.
(Choose $L_1=(a_1, b_1)=J_{k_1}$ so $0\in L_1$. Now there is exactly one $L_2=(a_2,b_2)=J_{k_2}$ with $b_1\in L_2$. Etc. Now, for example, $L_1\cap L_3=\emptyset$, since $x\in L_1\cap L_3$ implies that also  $x\in L_2$.)
