Prove that two paths on opposing corners of the unit square must cross.

I'm looking for a simple argument to the following:

Given two (continuous) paths on the unit square, one from (0,0) to (1,1) and the other from (1,0) to (0,1), prove that the paths cross at some point $(x_0, y_0)$.

I have constructed a topological argument for why this is true using compactness (and a proof by contradiction), but the person I'm working with seems to think there is a "simple" and "well-known" argument that says the two paths must cross. I haven't been able to find such a result. Does anyone know of one?

Thanks!

Here's an argument using the fundamental group. Suppose there were two such curves $a,b:[0,1]\to[0,1]^2$ that didn't meet anywhere, and say $a$ goes from $0,0)$ to $(1,1)$, while $b$ goes from $(0,1)$ to $(1,0)$. Consider the map $$c:[0,1]^2\to S^1,\;(s,t)\mapsto\frac{b(t)-a(s)}{|b(t)-a(s)|}$$ Now consider the restriction $c'$ of this map to the boundary $\partial[0,1]^2\simeq S^1$ which is made of four unit intervals (or paths) $(0,0)\xrightarrow{\;\alpha\;}(1,0)$, $(1,0)\xrightarrow{\;\beta\;}(1,1)$, $(1,1)\xrightarrow{\;\gamma\;}(0,1)$ and $(0,1)\xrightarrow{\;\delta\;}(0,0)$. Then $c\circ\alpha$ goes from $i$ to $-1$ while staying in the upper left quadrant of $\Bbb C\simeq \Bbb R^2$, $c\circ\beta$ goes from $-1$ to $-i$ while staying in the bottom left quadrant, $c\circ\gamma$ goes from $-i$ to $1$ while staying in the bottom right quadrant and $c\circ\delta$ goes from $1$ to $i$ while staying in the top right quadrant. This implies that $c'$ is homotopic to the identity map $S^1\to S^1$, which is a generator of $\pi_1(S^1)\simeq \Bbb Z$, yet by definition, it is also nulhomotopic, having an extension to the square, namely $c$. This yields a contradiction, and so the two paths $a$ and $b$ must meet.

I don't think there is a simple argument. This seems to be equivalent to the Jordan separation theorem, which, although not as difficult as the Jordan curve theorem, is still not exactly elementary.

The result is Lemma 2 of this paper by Maehara, which uses the Brouwer Fixed Point Theorem. (I'm not sure if this argument qualifies as "simple" since this theorem is usually proved using homology.) See also this MO thread.