gn.general topology - Pair of curves joining opposite corners of a square must intersect---proof?

Reposting something I posted a while back to Google Groups.

In his 'Ordinary Differential Equations' (sec. 1.2) V.I. Arnold says
"... every pair of curves in the square joining different pairs of
opposite corners must intersect".

This is obvious geometrically but I was wondering how one could go
about proving this rigorously. I have thought of a proof using
Brouwer's Fixed Point Theorem which I describe below. I would greatly
appreciate the group's
comments on whether this proof is right and if a simpler proof is
possible.

We take a square with side of length 1. Let the two curves be
$(x_1(t),y_1(t))$ and $(x_2(t),y_2(t))$ where the $x_i$ and $y_i$ are
continuous functions from $[0,1]$ to $[0,1]$. The condition that the
curves join different pairs of opposite corners implies,
$$(x_1(0),y_1(0)) = (0,0)$$
$$(x_2(0),y_2(0)) = (1,0)$$
$$(x_1(1),y_1(1)) = (1,1)$$
$$(x_2(1),y_2(1)) = (0,1)$$

The two curves will intersect if there are numbers $a$ and $b$ in $[0,1]$
such that

$$p(a,b) = x_2(b) - x_1(a) = 0$$
$$q(a,b) = y_1(a) - y_2(b) = 0$$

We define the two functions

$$f(a,b) = a + p(a,b)/2 + |p(a,b)| (1/2 - a)$$
$$g(a,b) = b + q(a,b)/2 + |q(a,b)| (1/2 - b)$$

Then $(f,g)$ is a continuous function from $[0,1]times [0,1]$ into itself and
hence must have a fixed point by Brouwer's Fixed Point Theorem. But at
a fixed point of $(f,g)$ it must be the case that $p(a,b)=0$ and $q(a,b)=0$
so the two curves intersect.

Figuring out what $f$ and $g$ to use and checking the conditions in the
last para is a tedious. Can there be a simpler proof?