The standard proof of Hopf's Umlaufsatz proceeds something like this: We have a unit speed $\mathcal{C}^1$ curve $\beta:\mathbb{R}\to\mathbb{R}^2$. Furthermore, $\beta$ is a simple loop with period $L$. We now define $T=\{(t_1,t_2) \in \mathbb{R}^2 : 0 \leq t_1 \leq t_2 \leq L \}$ and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) = \begin{cases} \beta'(t_1) & t_1=t_2 \\ -\beta'(0) & (t_1,t_2)=(0,L) \\ \frac{\beta(t_2)-\beta(t_1)}{\|\beta(t_2)-\beta(t_1)\|} & \text{otherwise} \end{cases} $$
It is visually obvious that $f$ is continuous, but I haven't found a strict proof of this anywhere. Below is my attempt at a proof, but I think it's ugly and long-winded and it also isn't complete. My question is how to finish it. I also suspect (or rather hope) that there is a significantly shorter (or more elegant) proof that is nevertheless complete and doesn't use hand-waving.
First of all, $f$ is continuous for all $(t_1,t_2)\in T$ with $t_1\neq t_2$ and $(t_1,t_2)\neq(0,L)$ because (as $\beta$ is simple) the denominator of $(\beta(t_2)-\beta(t_1))/(\|\beta(t_2)-\beta(t_1)\|)$ doesn't vanish and $\beta$ and the norm are continuous.
Now for the case $t_1=t_2$. Fix $t_1\in[0,L]$. We have $$ \lim_{t_2\to t_1} \frac{\beta(t_2)-\beta(t_1)}{t_2-t_1} = \beta'(t_1) $$ by definition. This implies $$ \lim_{\substack{t_2\to t_1\\t_2>t_1}} \frac{\|\beta(t_2)-\beta(t_1)\|}{t_2-t_1} = \lim_{\substack{t_2\to t_1}} \left\|\frac{\beta(t_2)-\beta(t_1)}{t_2-t_1}\right\| = \|\beta'(t_1)\| = 1 $$ as $\beta$ is a unit speed curve. Combining these two we get $$ \lim_{\substack{t_2\to t_1\\t_2>t_1}} f(t_1,t_2) = \lim_{t_2\to t_1} \frac{\beta(t_2)-\beta(t_1)}{t_2-t_1} \cdot \lim_{\substack{t_2\to t_1\\t_2>t_1}} \frac{t_2-t_1}{\|\beta(t_2)-\beta(t_1)\|} = \beta'(t_1) $$
This means that for every $\varepsilon>0$ we can find a $\delta>0$ such that for all $t_2$ with $|t_2-t_1|<\delta$ and $(t_1,t_2)\in T$ we have $\|f(t_1,t_2)-\beta'(t_1)\|<\varepsilon$.
As $[0,L]$ is compact, we can even, for a given $\varepsilon>0$, find a $\delta_1>0$ such that for all $(t_1,t_2)\in T$ with $|t_2-t_1|<\delta_1$ the inequality $\|f(t_1,t_2)-\beta'(t_1)\|<\varepsilon/2$ holds.
Also, as $\beta'$ is continuous and $[0,L]$ is compact, we can find a $\delta_2>0$ such that for all $t_1^\ast,t_1\in[0,L]$ with $|t_1-t_1^\ast|<\delta_2$ we have $\|\beta'(t_1) - \beta'(t_1^\ast)\|<\varepsilon/2$.
Now let $P=(t_1,t_2)\in T$ be arbitrary with $\|P-(t_1^\ast,t_1^\ast)\|<\min\{\delta_1/2,\delta_2\}$. That implies $|t_1-t_1^\ast|<\delta_2$ and $|t_2-t_1|<\delta_1$. By combining the previous two inequalities we get $\|f(P)-\beta'(t_1^\ast)\|<\varepsilon$. We have thus proved that $f$ is continuous in $(t_1^\ast,t_1^\ast)$.
The third case is the point $Q=(0,L)$. As $\beta$ is a loop, we have: $$ \lim_{t\to L} \frac{\beta(t)-\beta(0)}{L-t} = -\lim_{t\to L} \frac{\beta(L)-\beta(t)}{L-t} = -\beta'(L) = -\beta'(0) $$ And as a consequence we have: $$ \lim_{\substack{t\to L\\t<L}} \frac{\|\beta(t)-\beta(0)\|}{L-t} = \lim_{\substack{t\to L}} \left\|\frac{\beta(t)-\beta(0)}{L-t}\right\| = \|-\beta'(0)\| = 1 $$ Combining these two we get, as above: $$ \lim_{\substack{t\to L\\t<L}} f(0,t) = -\beta'(0) $$
At this point I'm stuck. I think that we need to show that $f$ is uniformly continuous around $Q$ and that maybe this is the case because $\beta$ has unit speed. If we can prove that, we can approach $Q$ in a way similar to the second case: starting from a point near enough we first move - by virtue of uniform continuity - parallel to the $x$ axis until the first component is zero. Then we move vertically towards $Q$.
EDIT:
Using Ted Shifrin's advice, I've rewritten the second case of the proof. However, it seems to me that this depends on the Taylor remainder $R$ being continuous in both variables (see question mark below) which - if justified in detail - is not much different from what I did above and still doesn't solve the third case. Or am I missing something?
As $\beta$ is differentiable, we can write, for each $t_1\in[0,L]$, $$ \beta(t_2)=\beta(t_1)+(t_2-t_1)\cdot\beta'(t_1) +(t_2-t_1)\cdot R(t_1,t_2-t_1) $$ with $\lim_{t\to0}R(t_1,t)=\mathbf0$.
We thus have $$ \lim_{(t_1,t_2)\to(t_1^\ast,t_1^\ast)} \frac{\beta(t_2)-\beta(t_1)}{t_2-t_1} = \lim_{(t_1,t_2)\to(t_1^\ast,t_1^\ast)} (\beta'(t_1) + R(t_1,t_2-t_1)) \stackrel{\color{red}?}{=} \beta'(t_1^\ast) $$ and $$ \lim_{\substack{(t_1,t_2)\to(t_1^\ast,t_1^\ast)\\t_2>t_1}} \frac{\|\beta(t_2)-\beta(t_1)\|}{t_2-t_1} = \lim_{\substack{(t_1,t_2)\to(t_1^\ast,t_1^\ast)\\t_2>t_1}} \left\|\frac{\beta(t_2)-\beta(t_1)}{t_2-t_1}\right\| = \|\beta'(t_1^\ast)\|=1 $$ which eventually leads to $$ \lim_{\substack{(t_1,t_2)\to(t_1^\ast,t_1^\ast)\\t_2>t_1}} f(t_1,t_2) = \beta'(t_1^\ast) $$ as above.
For the sake of completeness, I'll put the whole proof here. The "trick" is to enlarge the domain of $f$ so that both special cases are essentially the same except for the sign.
We have a unit speed $\mathcal{C}^1$ curve $\beta$. Furthermore, $\beta$ is a simple loop with period $L$. We now define $T=\{(t_1,t_2) \in \mathbb{R}^2 : 0 \leq t_1 \leq L \text{ and } t_1 \leq t_2 \leq t_1+L \}$ and a function $f$ on $T$ as follows: $$ f(t_1,t_2) = \begin{cases} \beta'(t_1) & t_2=t_1 \\ -\beta'(t_1) & t_2=t_1+L \\ \frac{\beta(t_2)-\beta(t_1)}{\|\beta(t_2)-\beta(t_1)\|} & \text{otherwise} \end{cases} $$ We want to prove that $f$ is continuous on $T$.
First of all, $f$ is continuous for all $(t_1,t_2)\in T$ with $t_2\notin\{t_1,t_1+L\}$ (the "otherwise" case in the definition) because (as $\beta$ is simple and has period $L$) the denominator of $(\beta(t_2)-\beta(t_1))/(\|\beta(t_2)-\beta(t_1)\|)$ doesn't vanish and $\beta$ and the norm are continuous.
Now for the case $t_2=t_1$. Fix $t_1\in[0,L]$. We have $$ \lim_{t_2\to t_1} \frac{\beta(t_2)-\beta(t_1)}{t_2-t_1} = \beta'(t_1) $$ by definition. This implies $$ \lim_{\substack{t_2\to t_1\\t_2>t_1}} \frac{\|\beta(t_2)-\beta(t_1)\|}{t_2-t_1} = \lim_{\substack{t_2\to t_1}} \left\|\frac{\beta(t_2)-\beta(t_1)}{t_2-t_1}\right\| = \|\beta'(t_1)\| = 1 $$ as $\beta$ is a unit speed curve. Combining these two we get $$ \lim_{\substack{t_2\to t_1\\t_2>t_1}} f(t_1,t_2) = \lim_{t_2\to t_1} \frac{\beta(t_2)-\beta(t_1)}{t_2-t_1} \cdot \lim_{\substack{t_2\to t_1\\t_2>t_1}} \frac{t_2-t_1}{\|\beta(t_2)-\beta(t_1)\|} = \beta'(t_1) $$
This means that for every $\varepsilon>0$ we can find a $\delta>0$ such that for all $t_2$ with $|t_2-t_1|<\delta$ and $(t_1,t_2)\in T$ we have $\|f(t_1,t_2)-\beta'(t_1)\|<\varepsilon$.
As $[0,L]$ is compact, we can even, for a given $\varepsilon>0$, find a $\delta_1>0$ such that for \textit{all} $(t_1,t_2)\in T$ with $|t_2-t_1|<\delta_1$ the inequality $\|f(t_1,t_2)-\beta'(t_1)\|<\varepsilon/2$ holds.
Also, as $\beta'$ is continuous and $[0,L]$ is compact, we can find a $\delta_2>0$ such that for all $t_1^\ast,t_1\in[0,L]$ with $|t_1-t_1^\ast|<\delta_2$ we have $\|\beta'(t_1) - \beta'(t_1^\ast)\|<\varepsilon/2$.
Now let $P=(t_1,t_2)\in T$ be arbitrary with $\|P-(t_1^\ast,t_1^\ast)\|<\min\{\delta_1/2,\delta_2\}$. That implies $|t_1-t_1^\ast|<\delta_2$ and $|t_2-t_1|<\delta_1$. By combining the previous two inequalities we get $\|f(P)-\beta'(t_1^\ast)\|<\varepsilon$. We have thus proved that $f$ is continuous in $(t_1^\ast,t_1^\ast)$.
The third case is $t_2=t_1+L$. This is almost identical to the second case. He we have $$ \lim_{t_2\to t_1+L} \frac{\beta(t_2)-\beta(t_1+L)}{(t_1+L)-t_2} = -\beta'(t_1+L) = -\beta'(t_1) $$ because $\beta$ has period $L$. And thus: $$ \lim_{\substack{t_2\to t_1+L\\t_2<t_1+L}} \frac{\|\beta(t_2)-\beta(t_1+L)\|}{(t_1+L)-t_2} = \lim_{\substack{t_2\to t_1+L}} \left\|\frac{\beta(t_2)-\beta(t_1+L)}{(t_1+L)-t_2}\right\| = 1 $$ When we combine these two, the only difference is the sign: $$ \lim_{\substack{t_2\to t_1+L\\t_2<t_1+L}} f(t_1+L,t_2) = -\beta'(t_1) $$ We can now proceed like above to show that $f$ is continuous for all points of the form $(t_1^\ast,t_1^\ast+L)$ and thus in particular at $(0,L)$.