Here is the proof of the equivalent statement "Every complex non-constant polynomial $p$ is surjective".
1) Let C be the finite set of critical points ($f'(z)=0$). C is finite by elementary algebra.
2) Remove from the codomain $p(C)$ (and call the resulting open set B) and from the domain its inverse image (again finite) (and call the resulting open set A).
3) Now you get an open map from A to B, which is also closed, because any polynomial is proper (inverse images of compact sets are compact). But B is connected and so $p$ is surjective.
I like this proof because you can try it for real polynomials and it breaks down at step 3) because if you remove a single point from the line you disconnect it, while you can remove a finite set from a plane leaving it connected.
