The following is a sketch of Hopf's own proof:
Assume that $\gamma$ is given in the form $\gamma:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$ with ${\bf z}'(t)\ne {\bf 0}\quad\forall t,\quad {\bf z}(0)=(0,0),\quad {\bf z}'(0)=(1,0), \quad y(t)\geq 0\quad\forall t\ .$

Consider the triangular domain $B:=\bigl\{(s,t)\ \bigm|\ 0\leq s\leq t\leq T\bigr\}$ and on $B$ the function ${\bf g}(s,t):=\int_0^1{\bf z}'\bigl((1-\tau) s +\tau t\bigr)\ d\tau =\cases{{{\bf z}(t)-{\bf z}(s)\over t-s} \quad &$(s From the assumptions on $\gamma$ it follows that ${\bf g}(s,t)\ne{\bf 0}$ on $B$. Since $B$ is simply connected the function ${\bf g}$ therefore possesses a continuous real-valued argument $\theta(\cdot,\cdot)$ on $B$, i.e., there exists a continuous function $\theta:\ B\to{\mathbb R}$ such that $\bigl[\theta(s,t)\bigr]=\arg{\bf g}(s,t)\qquad\bigl((s,t\in B\bigr)\ .$ We may assume $\theta(0,0)=0$. Since $\bigl[\theta(s,s)]=\arg{\bf z}'(s)$ we may consider $\hat\theta(s):=\theta(s,s)$ as a continuous real-valued argument of ${\bf z}'(s)$ along $\gamma$. It remains to prove that $\hat\theta(T)=\theta(T,T)=2\pi$.
To this end we look at the behavior of $\theta(0,t)=\arg\bigl({\bf z}(t)-{\bf z}(0)\bigr)\qquad(0\leq t\leq T)$ along the vertical edge of $B$. The limiting direction at $(0,T)$ is parallel to $(-1,0)$, and as $\arg\bigl({\bf z}(t)-{\bf z}(0)\bigr)\ne-{\pi\over2}$ for all $t\in[0,T]$ it follows that necessarily $\theta(0,T)=\pi$. Similarly we then look at the behavior of $\theta(s,T)=\arg\bigl({\bf z}(T)-{\bf z}(s)\bigr)\qquad(0\leq s\leq T)$ along the horizontal edge of $B$. The limiting direction at $(T,T)$ is parallel to $(1,0)$, and as $\arg\bigl({\bf z}(T)-{\bf z}(s)\bigr)\ne{\pi\over2}$ for all $s\in[0,T]$ it follows that necessarily $\theta(s,s)=2\pi$.