The proposition is as follows:
Let segment $AB$ be given and the points $C_0=A$, $C_1$, $\ldots$, $C_n=B$ and $D_0=A$, $D_1$, $\ldots$, $D_n=B$ be on the same side of the line such that the broken line $C_0C_1\cdots C_n$ is inside the region bound by $D_0D_1\cdots D_n$ and the segment $AB$. Then the length of $C_0C_1\ldots C_n$ is smaller than that of $D_0D_1\cdots D_n$.
This is just my speculation and I couldn't either prove it or disprove it. The proposition came to my mind after studying the proof of the iterative version of the triangle inequality for a broken segment; I know that any broken line is longer than the straight segment $AB$, and now I want to compare the lengths between the broken segments. I thought about using a different index for $C_i$ up to $C_m$ with $m\neq n$, but then I can just insert arbitrarily many lines within the bouned region so it seemed like it wouldn't work. I'd be very grateful if any other generalization in this direction is provided as well.
Your conjecture is false.
Let $A = C_0 = D_0 = (0, 0)$, $C_1 = D_1 = (0, 1)$, $D_2 = (1, 1)$, $C_3 = D_3 = (2, 1)$, and $B = C_4 = D_4 = (2,0)$; and $C_2 = (1, 0)$. Then $\overline{C_1 C_2} = \overline{C_2 C_3} > \overline{D_1 D_2} = \overline{D_2 D_3}$, all other line segments have the same length, and the $C_n$ are all inside the $D_n$. (If you need them to be strictly inside, add or subtract small $\epsilon$ from the $C_n$ coordinates as needed.)
EDIT: It does hold for $n = 2$,though. If $\overline{D_0 D_1 D_2}$ is collinear, then so is $\overline{C_0 C_1 C_2}$. And if not, then the set of points $P$ where $\overline{A P} + \overline{P B}= \overline{A D_1} + \overline{D_1 B}$ is an ellipse going through $D_1$. $C_1$ is inside the triangle, therefore it's inside the ellipse, and therefore its length can't be larger.