Assume $f:\mathbb{R}^n \to \mathbb{R}$ is convex, and $L$-Lipschitz, so $|f(x)-f(y)|\leq L\|x-y\|$. I would like to show that $\|\nabla f(x)\|\leq L$.
In one dimension this is a straightforward consequence of the fact that convexity implies $f(y)-f(x)\geq f'(x) (y-x), \forall x, y \in \mathbb{R}$, but I'm having trouble translating this to several variables (in particular, Cauchy-Schwarz is working on the opposite direction!). If it helps, I don't mind assuming the domain of $f$ is a compact $[a,b]^n$.
In $\mathbb{R}^n$, we have that a function is convex iff $$f(\mathbf{y})-f(\mathbf{x}) \geq Df(\mathbf{x})(\mathbf{y}-\mathbf{x})$$ Where $Df(\cdot)$ is the Jacobian of $f$.
For a proof, see, for example, "Mathematics for Economists" by Simon & Blume, page 511.