Suppose we have a continous curve $\gamma:[a,b]\rightarrow\mathbb{R}^2$ where all one-sided derivatives exist. By $\gamma_\epsilon:[a,b]\rightarrow\mathbb{R}^2$ we mean the outside shifted parallel curve, such that euclidean distance $\epsilon$ between the images in $\mathbb{R}^2$ is preserved, see https://en.wikipedia.org/wiki/Parallel_curve#Parallel_curve_to_a_curve_with_a_corner. To be exact (since the following parametrization differs from the one in the first paragraph of the wikipedia article): having the image of the outwards shifted parallel curve, we set $\gamma_\epsilon(t):=\lambda(t)\gamma(t)$, such that $\lambda\geq 1$ regulates the distance in a way, that $\gamma_\epsilon(t)$ indeed lies on the image of the parallel curve.
Now it seems senseful that $\gamma_\epsilon\rightarrow \gamma$ uniformly, which is to be shown.
What did I try? After drawing some examples (with corners), I suppose we get something like $\lambda\leq 1+\epsilon$ which would be sufficient due to $\|\gamma\|$ being finite. However, I could not deduce that analytically. Another idea I had was to view the problem geometrically by applying the Law of Cosines. I set $t_\epsilon$ such that $|\gamma(t_\epsilon)-\gamma_\epsilon(t)|=\epsilon$ and looked at the triangle with corners $\gamma(t),\gamma_\epsilon(t),\gamma(t_\epsilon)$ where we can at least bound the length of one side. I could not conclude further information, though. Lastly, I thought of applying Dini's Theorem to $\mu_\epsilon(t):=|\gamma_\epsilon(t)-\gamma(t)|$ which seemed promising since pointwise convergence and monotonicity both seem obvious to my eye, but explicitly showing both is another story I could not tell.
EDIT If necessary, you might also consider a different parametrization. I just thought the mentioned one does the job. If a different one is better suitable, that would help a lot, too.