As far as I'm aware, there isn't a compelling direct argument from Brouwer fixed point to imply the fundamental theorem of algebra. Such an argument isn't impossible -- I can imagine some fairly contrived proofs but I don't know of a very natural one. The references like Guillemin and Pollack don't derive FTOA from Brouwer, they derive both FTOA and Brouwer from degree/intersection theory. In particular they only use mod-2 degree theory for Brouwer but oriented degree theory for FTOA.

I had an argument written down here previously that I thought might work but now I realize it can't work. Oh, but it's fixable.

EDIT *I've managed to repair the argument*. The downside is it's not as simple.

A polynomial without roots produces a polynomial without fixed points. Specifically, $p(z) \neq 0$ for all $z \in \mathbb C$ means $q(z) = p(z)+z$ has no fixed points in $\mathbb C$. So what? Think of $q(z)$ as a map of the Riemann sphere. Now take the real oriented blow-up of the Riemann sphere at infinity (i.e. replace the point at infinity by its unit normal bundle in the sense of smooth real manifolds). This is a disc. So $q(z)$ becomes a smooth map of the disc, denote it $\hat q$, and identify the blow-up with $D^2$, the unit disc in $\mathbb C$ centred at the origin.

If $q(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$ and if we conjugate by $z \longmapsto 1/z$ so

$$\frac{1}{q(1/z)} = \frac{z^n}{1+a_{n-1}z + \cdots + a_0z^n}$$

So when you restrict $\hat q$ to the boundary circle, it becomes $z \longmapsto z^n$.

$z \longmapsto z^n$ has fixed points $z^{n-1}=1$, the $(n-1)$-th roots of unity. So we can not directly appeal to Brouwer, since Brouwer's fixed point theorem might give you a pre-existing fixed point on the boundary.

Consider the vector field $v(z) = z-\hat q(z)$ on $D^2$. It is inward-pointing on the boundary circle with the sole exception of $z^{n-1}=1$, the $(n-1)$ roots of unity. But if we remove a small neighbourhood of $\partial D^2$ from $D^2$, the vector field $v$ restricts to an *inward pointing* vector field. So you could appeal to Poincare-Hopf and say there has to be a zero in the interior, or you could talk about the flow of the vector field, and Brouwer's fixed-point theorem would then tell you the vector field must have a zero in the interior.

So its not a slick proof, but it can be done.

A suitable identification between $D^2$ and the blow-up of the Riemann sphere at infinity is done by the map $X : D^2 \to \hat{\mathbb C}$ given by $X(z) = \frac{1}{1-|z|^2} z$. So $\hat q$ is the unique continuous extension of $X \circ q \circ X^{-1}$ to $D^2$.