Let $x,y\in\mathbb{R}^n\setminus\{0\}$. Define $\mathrm{pol}:\mathbb{R}^n\setminus\{0\}\to\mathbb{S}^{n-1}$ by $\mathrm{pol}(z):=\frac{z}{\|z\|}\forall z\in\mathbb{R}^n\setminus\{0\}$.
Question Is the following claim correct, and if it is, can it be improved also for small $\|y\|,\|x-y\|$? It seems to me like the estimate should break down when approaching the origin, as $\mathrm{pol}$ is not continuously extendible there, but I just want to be sure I'm not missing a more obvious derivation. Very close to the origin, $\|x-y\|,\|y\|$ can get arbitrarily small yet $\|\mathrm{pol}(x)-\mathrm{pol}(y)\|$ can be held fixed (if we travel along the rays downwards toward the origin).
Claim We have for some $1<C<\infty$, $$\|\mathrm{pol}(x)-\mathrm{pol}(y)\|\leq C \frac{\|x-y\|}{1+\|y\|}$$ if I: $\|x-y\|<\|y\|$ and $\|y\|\geq(C-1)^{-1}$ or II: $\|x-y\|\geq\|y\|$ and $C\|x-y\|-1\geq\|y\|$.
Proof We start from the basic estimate which is $\|\mathrm{pol}(x)-\mathrm{pol}(y)\|\leq2$ for the worst case of two antipodal points. Furthermore we have \begin{align} |\mathrm{pol}(x)-\mathrm{pol}(y)| &= \frac{1}{\|y\|}\|\frac{\|y\|x-\|x\|y}{\|x\|}\| \\ &=\frac{1}{\|y\|}\|\frac{(\|y\|-\|x\|)x-\|x\|(y-x)}{\|x\|}\| \\ &\leq 2\frac{\|x-y\|}{\|y\|}\,. \end{align} This holds for all $x,y\neq0$. Combining these two we find $$ |\mathrm{pol}(x)-\mathrm{pol}(y)| \leq 2 \min(\{1,\frac{\|x-y\|}{\|y\|}\}) \,.$$ This results in the above claim by choosing any $C>1$.