The mapping $f:[-1,1]^2\to\mathbb R^2$ is known to be continuous. Also the image of the upper edge of the rectangle is contained in the upper half-plane, the left edge's image is contained in the left half-plane, and so on for the bottom and right edge. Formally speaking this means:
If $f_1,f_2:[-1,1]^2\to\mathbb R$ are defined to be the components of $f$, i. e. $$f(x,y)=(f_1(x,y),f_2(x,y))\,,$$ then the following conditions hold
- for each $x\in[-1,1]$ it holds $f_2(x,1)>0$,
- for each $y\in[-1,1]$ it holds $f_1(-1,y)<0$,
- for each $x\in[-1,1]$ it holds $f_2(x,-1)<0$ and
- for each $y\in[-1,1]$ it holds $f_1(1,y)>0$.
Under these assumptions prove that there exist $x_0,y_0\in[-1,1]$ such that $f(x_0,y_0)=(0,0)$. This appears to be obvious intuitively, but I'm looking for a formal and rigorous proof.