A visually guided proof of the fundamental theorem of algebra?

A complex root of a polynomial $P(z)$ is a pair of real numbers $u,v$ that simultaneously make the real part and the imaginary part of $P(z)$ zero.

The zeros of the real part and the imaginary part are given by two curves in the complex plane

$$\operatorname{Re}(P(u +iv)) = 0$$

$$\operatorname{Im}(P(u+iv)) = 0$$

The zeros of the polynomial are the intersection points of these two curves.

For the sake of specifity here are the two curves for $P(z) = z^5 + z^3 + z^2 + z + 1$:

$u^5-6u^3v^2-4v^2u^3+5uv^4+ u^3-v^2u-2uv^2+u^2-v^2+u+1 = 0 $

$v^5+5vu^4-10u^2v^3+3u^2v-v^3+2uv+v=0$

The fundamental theorem of algebra says that such curves always do intersect. A proof of the fundamental theorem might go like this: The curves $\operatorname{Re}(P(z)) = 0$ (red) and $\operatorname{Im}(P(z)) = 0$ (blue) – which are tightly related – are such-and-such, so they must intersect at least once and at most $n$ times (for $n$ the degree of the polynomial).

What can be seen is, that the curves always come in $n$ branches which extend to infinity and for some reason must intersect.

$x^2 + x + 1$

$x^3 + x^2 + x + 1$

$x^4 + x^2 + x + 1$

$x^5 + x^3 + x^2 + x + 1$

How could such a proof be spelled out?

That idea is basically the approach to the Fundamental Theorem of Algebra taken by Gauss, in his PhD thesis Demonstratio nova theorematis omnem functionem algebraicam rationalem integram unius variabilis in factores reales primi vel secundi gradus resolvi posse. It's a nice geometrical approach to the theorem, but hard to complete rigorously. I suggest that you read C. F. Gauss’s proofs of the Fundamental Theorem of Algebra, by Harel Cain.