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 $b_j$.
Best Answer
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.