Why there is no continuous argument function on $\mathbb{C}\setminus\{0\}$?

An argument function $\phi$ on $\mathbb{C}\setminus\{0\} = \mathbb{R}^2\setminus\{0\}$ is a function such that for every $z\neq 0$ it holds that $$z = |z|\exp(i\phi(z)).$$

Is there an elementary and easy proof that there is no continuous argument function on $\mathbb{C}\setminus\{0\}$? I would like to see a proof which uses as less complex analysis as possible. Probably only topological arguments and no complex numbers whatoever?

If there were a continuous argument function, its restriction to the unit circle would be a homeomorphism onto its image in $\mathbb R$. (It is injective, the circle is compact, and $\mathbb R$ is Hausdorff.) But the image of the circle would be a compact and connected subset of $\mathbb R$, thus a closed and bounded interval, which is not homeomorphic to the circle. This contradiction shows that such a function doesn't exist.

Since the question itself uses complex numbers, I don't think it's possible to give a proof that doesn't use them.

Anyway, the following is a simple argument that doesn't use any complex analysis.

Suppose such a function $\phi$ exists, and consider the function $\psi(t) = \frac{1}{2 \pi}(\phi(e^{i t}) - t)$, for real $t$. $\psi$ is continuous and $\mathbb{Z}$-valued, so it must be constant, say $\psi \equiv k$.

So $\phi(e^{it}) = 2 \pi k + t$, but then

$$ 2 \pi k = \phi(1) = \phi\left(e^{2 \pi i}\right) = 2 \pi (k + 1), $$

contradiction.