Fix two points $0 \neq a,b \in \mathbb{R}^n$ with the euclidean metric and consider the function
$$ f: \mathbb{R}^n \rightarrow [0,\pi]; x \mapsto \sphericalangle (axb), $$
where $\sphericalangle (axb)$ denotes the angle at $x$ in the triangle $\triangle abx$.
Is this function Lipschitz continuous in a small neighborhood around $0$?
Let $L=\Vert a-b\Vert$ be the distance between $a$ and $b$, and let $\alpha(x)$ be the angle of the triangle $\triangle abx$ at $x$.
Some basic trigonometry yields $$\cos\alpha(x)=\frac{\Vert a-x\Vert^2+\Vert b-x\Vert^2-L^2}{2\Vert a-x\Vert\Vert b-x\Vert}$$
The functions involved are all differentiable at $x\neq a,b$, so $\cos\alpha$ is differentiable at all $x\neq a,b$.
Moreover, $\alpha=\arccos \cos\alpha$, and $\arccos:[-1,1]\to[0,\pi]$ is differentiable on $(-1,1)$. Let's see when $\cos\alpha(x)\neq \pm 1$
We have \begin{align*} \cos\alpha x=\pm 1&\iff\Vert a-x\Vert^2+\Vert b-x\Vert^2-L^2=\pm 2\Vert a-x\Vert \Vert b-x\Vert\\ &\iff(\Vert a-x\Vert\pm\Vert b-x\Vert)^2=L^2\\ &\iff\Vert a-x\Vert\pm\Vert b-x\Vert=\pm L=\pm\Vert a-b\Vert \end{align*} which happens when $a,b$ and $x$ are colinear.
Thus, denoting by $\overline{ab}$ the line which passes through $a$ and $b$, we see that $\alpha$ is differentiable on $\mathbb{R}^n\setminus\overline{ab}$, and hence locally Lipschitz.
The case where $a,b,x$ are colinear seems a little more complicated.
Thus, as long as $a$ and $b$ are linearly independent, the angle map is Lipschitz on a small neighbourhood around $0$.