Sawtooth-like polygonal chains in $\mathbb R^2$ must self-intersect. We consider closed polygonal chains in the 2-dimensional plane with an even number of sides, say $2n$, numbered as $A_1B_1A_2B_2\dots A_nB_nE$, where $E = A_1$.
We require additionally, that each $B_i$ lies right below $A_i$, like with sawtooth.
Edit: Since right below seems ambiguous, let's state this line explicitly as that the $y$-coordinate of each $B_i$ be strictly less than that of $A_i$.
Is it true that any such polygonal chain must self-intersect?
In the 3-dimensional case, there is the obvious counterexample of traversing the vertices of any prism in the obvious order. I have however yet to find a proof or counterexample in the 2-dimensional case.

Yes. For simplicity let us assume that "polygon" means that consecutive edges cannot be parallel (no 180-degree angles at vertices). In particular this implies that the outgoing edges from $B_i$ are not vertical.
If the polygon does not intersect itself, it is oriented either clockwise or counterclockwise.
If clockwise, consider any leftmost vertex $v$, and the vertical line $L$ passing through it. No vertex of the polygon lies to the left of $L$. Because we do not allow 180-degree angles, and because of the sawtooth condition, exactly one of the edges incident at $v$ will be vertical, and it must be oriented up because the polygon is oriented clockwise. This is impossible, because all vertical edges must be oriented down by the sawtooth condition.
If the polygon is oriented counterclockwise, repeat the same reasoning at any rightmost vertex; because the polygon is oriented counterclockwise, the incident vertical edge must be oriented up, which is impossible.
