This feels as though it should be falsifiable, but it's not immediately obvious to me. The informal version of the statement is 'for every non-intersecting curve between two opposite corners of a square, there's a curve between the other two corners that only intersects it once.' Formally:
Let $f(): [0, 1]\mapsto [0,1]^2$ be a non-self-intersecting curve with $f(0) = \langle0,0\rangle$, $f(1) = \langle1,1\rangle$, and $f(t)\in (0,1)^2$ for $t\in(0,1)$. (Note that I'm requiring that all but the endpoints of the curve lie in the open square!) Then there exists a non-self-intersecting curve $g(): [0, 1]\mapsto [0,1]^2$ with $g(0) = \langle1,0\rangle$, $g(1) = \langle0,1\rangle$, and $g(t)\in (0,1)^2$ for $t\in(0,1)$ such that there are unique $t_0$ and $ t_1$ with $f(t_0) = g(t_1)$.
This feels like it ought to be a consequence of the JCT and/or Schoenflies' theorem, but the catch is that I don't see any clean ways of ensuring that a $g()$ constructed by those theorems actually maps back to the interior of the square as opposed to its boundary.
This is either right or, you know, just wrong. But here goes:
Let $L,R,T,B$ be the left, right, top, and bottom edges of the square, and let $F$ be the image of $f$. $C_{1}=F\cup R\cup B$ is a simple closed curve, so by Jordan–Schönflies, $C_{1}$ together with the region $R_{1}$ that it encloses is homeomorphic to a closed disk $D_1$. The homeomorphism can be chosen so that $(1,0)$ is mapped to the circle's north pole while $F$ is mapped onto the lower semicircle.
Similarly, $C_{2}=F\cup T\cup L$ together with the region $R_{2}$ that it encloses is homeomorphic to a closed disk $D_2$, where $(0,1)$ is mapped to the north pole and $F$ to the lower semicircle. This homeomorphsim can be further chosen so that the point on $F$ mapped to the south pole of $D_2$ is the same as the point on $F$ that was mapped to the south pole of $D_1$.
For $i=1,2$, let $V_i$ be the vertical diameter of $D_i$, and let $G_i$ be the image of $D_i$ under the homeomorphism sending $D_i$ to $C_i\cup R_i$. Then the concatenation $G=G_1\cup G_2$ is the desired non-self-intersecting curve.