Suppose we have two curves, $A$ and $B$. The domain of $A$ is $D_x(A)=\{x_p|P\in A\}=(x_1;x_2),D_y(A)=\{y_p|P\in A\}=(y_1;y_2),\frac{y_2-y_1}{x_2-x_1}\ll 1$. Similarly, $D_x(B)=\{x_p|P\in B\}=(x_1;x_2),D_y(B)=\{y_p|P\in B\}=(y_1;y_2),\frac{x_2-x_1}{y_2-y_1}\ll 1$. (Roughly, if you zoom far enough out, they form a cross). Neither $A$ nor $B$ is required to be "nice" - they must merely be continuous, but it is not required that they be differentiable.
Then $A$ undergoes a continuous translation whose total length is much less than the length of either of the curves. Let $S$ be the set of points which are in $A\cap B$ at any point during the translation.
Is it true that there is always a point $p_0\in A\cap B$ at time $t=0$ and a point $p_1\in A\cap B$ at time $t=T$ such that there is a connected component in $S$ which includes both points? (The choice of points is made freely; only one such pair need exist)
It seems obvious to me that this is true, but I cannot prove it. Given the difficulty of the Jordan Curve theorem (and my lack of a background in geometry), I don't expect it to be simple. Apologies for any lack of rigor; I hope I've communicated it clearly.