Is the function $x\cdot|x| \equiv \text{sgn}(x)\cdot x^2$ Locally-Lipschitz?
I believe is indeed locally Lipschitz since is continuous ans its slope is bounded on every compact-supported section.... but since $x\cdot|x| \equiv x\sqrt{x^2}$ and $\sqrt{x}$ is not locally Lipschitz, I am not $100\%$ sure... hope you can corroborate it.
Consider $[-a,a]$. Take $C=2a$. Then, $|sgn (x)x^{2}-sgn (y)y^{2}| \leq C|x-y|$ is clear if $x$ and $y$ have the same sign. Suppose $x$ is positive and $y$ is negative. Then $x^{2}+y^{2} \leq ax+a|y|=a(x-y)=a|x-y|\leq C|x-y|$. A similar argument holds if $y$ is positive and $x$ is negative.