$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}$Not sure this is useful, but a few years ago I put some thought into what it would take to rigorously give this proof in a Honors Multivariable calculus course (the sort where everything is proved). Here is what I came up with. Let $f:\CC \to \CC$ be a polynomial of degree $n$. It turns out to be convenient to make the following simplifying assumption -- letting $w_1$, $w_2$, ..., $w_k$ be the zeroes of $f'$, none of $f(w_i)$ are either purely real or purely imaginary. This is possible because, if $f(w_i)=0$ we are done, and otherwise we can replace $f$ by $e^{i \theta} f$ for some generic $\theta$.

Let $R = f^{-1}(\RR)$ and $I = f^{-1}(i \RR)$. The argument which we are trying to make rigorous is "near infinity, $R$ and $I$ look like $4n$ interleaved spokes, so they must cross somewhere in the interior." The implicit function theorem shows that $I$ and $R$ are closed one dimensional submanifolds of $\CC$. (This is why we required that the zeroes of $f'$ be disjoint from $I \cup R$.)

From this perspective, we can see that it would be bad if one of the components of $R$, for example, stopped at a point, or spiralled infinitely into a point -- we need them to disconnect $\mathbb{C}$.

I have heard people say that fixing this proof comes down to proving the Jordan curve theorem in the form "If $\phi: \RR \to \CC$ and $\psi: \RR \to \CC$ are smooth maps which are interleaved at infinity, then they $\phi(\RR)$ and $\psi(\RR)$ cross. In fact, I claim the hard thing is to show that the unbounded components of $R$ may be parametrized by $\RR$ in the first place, and that each component has two ends!

First, let's see why things are easy if we assume such a parametrization exists. Let $\Gamma$ be a connected component of $R$ touching one of the unbounded spokes. Suppose we could show there is a parametrization $\phi: \RR \to \Gamma$. Note that the composite $f \circ \phi: \RR \to \RR$ has nowhere vanishing derivative, so it is monotone and without loss of generality we can assume it is increasing. If we know that $\Gamma$ is unbounded in both directions (which is what is implicitly assumed when you draw a picture of $R$ as a bunch of strands connecting the spokes at infinity), then there is no need to use the Jordan curve theorem -- just apply the implicit function theorem to $f \circ \phi$! We have $\lim_{t \to \pm \infty} f(\phi(t))=\pm \infty$, so somewhere in the middle $f(\phi(t))=0$ and we win.

So the real challenge is to show that $\Gamma$, a connected unbounded $1$-dimensional submanifold of $\CC$, can be parametrized by $\RR$ and goes to $\infty$ in both directions. I looked up various proofs of the classification of $1$-dimensional manifolds, but they all seemed a little messier than I wanted to do in class.

Then I came up with the idea of just trying to invert the map $f: \Gamma \to \RR$, which is how I came up with this argument. I must admit, though, that the geometric origins are no longer visible.

$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}$Not sure this is useful, but a few years ago I put some thought into what it would take to rigorously give this proof in a Honors Multivariable calculus course (the sort where everything is proved). Here is what I came up with. Let $f:\CC \to \CC$ be a polynomial of degree $n$. It turns out to be convenient to make the following simplifying assumption -- letting $w_1$, $w_2$, ..., $w_k$ be the zeroes of $f'$, none of $f(w_i)$ are either purely real or purely imaginary. This is possible because, if $f(w_i)=0$ we are done, and otherwise we can replace $f$ by $e^{i \theta} f$ for some generic $\theta$.

Let $R = f^{-1}(\RR)$ and $I = f^{-1}(i \RR)$. The argument which we are trying to make rigorous is "near infinity, $R$ and $I$ look like $4n$ interleaved spokes, so they must cross somewhere in the interior." The implicit function theorem shows that $I$ and $R$ are closed one dimensional submanifolds of $\CC$. (This is why we required that the zeroes of $f'$ be disjoint from $I \cup R$.)

From this perspective, we can see that it would be bad if one of the components of $R$, for example, stopped at a point, or spiralled infinitely into a point -- we need them to disconnect $\mathbb{C}$.

I have heard people say that fixing this proof comes down to proving the Jordan curve theorem in the form "If $\phi: \RR \to \CC$ and $\psi: \RR \to \CC$ are smooth maps which are interleaved at infinity, then they $\phi(\RR)$ and $\psi(\RR)$ cross. In fact, I claim the hard thing is to show that the unbounded components of $R$ may be parametrized by $\RR$ in the first place, and that each component has two ends!

First, let's see why things are easy if we assume such a parametrization exists. Let $\Gamma$ be a connected component of $R$ touching one of the unbounded spokes. Suppose we could show there is a parametrization $\phi: \RR \to \Gamma$. Note that the composite $f \circ \phi: \RR \to \RR$ has nowhere vanishing derivative, so it is monotone and without loss of generality we can assume it is increasing. If we know that $\Gamma$ is unbounded in both directions (which is what is implicitly assumed when you draw a picture of $R$ as a bunch of strands connecting the spokes at infinity), then there is no need to use the Jordan curve theorem -- just apply the implicit function theorem to $f \circ \phi$! We have $\lim_{t \to \pm \infty} f(\phi(t))=\pm \infty$, so somewhere in the middle $f(\phi(t))=0$ and we win.

So the real challenge is to show that $\Gamma$, a connected unbounded $1$-dimensional submanifold of $\CC$, can be parametrized by $\RR$ and goes to $\infty$ in both directions. I looked up various proofs of the classification of $1$-dimensional manifolds, but they all seemed a little messier than I wanted to do in class.

Then I came up with the idea of just trying to invert the map $f: \Gamma \to \RR$, which is how I came up with this argument. I must admit, though, that the geometric origins are no longer visible.