How to prove that continuous choice of argument exists? The following proposition is given without proof in these lecture notes:
Proposition 4.5.1.: Let $\gamma$ be a path in $\mathbb C \setminus \{0\}$. Then there exists a parametrisation $\gamma: [a,b]\to \mathbb C \setminus \{0\}$ of $\gamma$ for which $t \mapsto \arg \gamma (t)$ is a continuous function. 
I tried to find a proof of this in other lecture notes online but since it does not appear to have a name I could not find anything.


If $\gamma$ is a curve in $\mathbb C \setminus \{0\}$ not passing
   through the origin how can I prove that there exists a continuous map
   $\alpha$ (the argument) so that $\gamma(t) = |\gamma (t)|
 e^{i\alpha(t)}$?

 A: Since the trace of $\gamma$ is compact, there is an $r > 0$ such that $\lvert \gamma(t)\rvert \geqslant r$ for all $t\in [a,b]$. By uniform continuity of $\gamma$, there is an $\varepsilon > 0$ such that
$$\lvert t-s\rvert \leqslant \varepsilon \implies \lvert\gamma(t) - \gamma(s)\rvert \leqslant r.$$
Thus for every interval $I\subset [a,b]$ of length $\leqslant \varepsilon$, $\gamma(I)$ is contained in a domain on which a continuous branch of $\arg$ exists (and hence infinitely many continuous branches, all differing by multiples of $2\pi$), namely a half-plane. Pick an $n\in \mathbb{N}\setminus \{0\}$ such that $\frac{b-a}{n}\leqslant \varepsilon$, and let $I_k = \bigl[ a + (k-1)\frac{b-a}{n}, a + k\frac{b-a}{n}\bigr]$ for $1 \leqslant k \leqslant n$.
Choose an arbitrary branch of $\arg$ defined on a neighbourhood of $\gamma(I_1)$ and define $\alpha(t) = \arg \gamma(t)$ on $I_1$ with that branch. Once $\alpha$ is defined on $I_1 \cup \dotsc \cup I_k$, if $k < n$, on a neighbourhood of $\gamma(I_{k+1})$ choose the branch $\arg_{k+1}$ of $\arg$ with $\arg_{k+1} \gamma\bigl(a + k\frac{b-a}{n}\bigr) = \alpha\bigl(a + k\frac{b-a}{n}\bigr)$, and define $\alpha(t) = \arg_{k+1} \gamma(t)$ for $t\in I_{k+1}$. The choice of $\arg_{k+1}$ makes $\alpha$ well-defined and continuous on $I_1 \cup \dotsc \cup I_k \cup I_{k+1}$. After $n-1$ steps, $\alpha$ is defined on all of $[a,b]$.

If there is some topological machinery available, one can avoid an explicit construction, and note that since the interval $[a,b]$ is simply connected, and $\exp \colon \mathbb{C} \to \mathbb{C}\setminus \{0\}$ is the universal covering, there exists a lift $\tilde{\gamma} \colon [a,b] \to \mathbb{C}$ of $\gamma$ - that is, $\exp \circ \tilde{\gamma} = \gamma$ - and one can define $\alpha(t) = \operatorname{Im} \tilde{\gamma}(t)$.
