Locally continuously differentiable implies locally Lipschitz
Yes, it is still true. Given $x_0$, use the continuity of $f'$ to find a neighborhood $U$ of $x_0$ such that $\|f'(x)-f'(x_0)\|\le 1$ for all $x\in U$. It follows that $\|f'(x)\|\le \|f'(x_0)\|+1$ in $U$. Now the Mean Value Theorem implies that $f$ is Lipschitz on $U$.