Finite convex functions are locally Lipschitz

Choose a closed cube centered at $x_0$ whose concentric dilation by factor $2$ lies in $U$. Every point of the cube is a convex combination of its finitely many vertices, so [F1] and convexity bound $f$ above there by the largest vertex value $M$. If $x$ lies in the cube, then $2x_0-x$ lies in the dilated cube; convexity gives $f(x_0)\le (f(x)+f(2x_0-x))\mathrm{/}2$, hence $f(x)\ge 2f(x_0)-M^{\mathrm{\prime }}$, where $M^{\mathrm{\prime }}$ is the largest value of $f$ at the finitely many vertices of the dilated cube, an upper bound there by the same convex-combination argument. Thus $f$ is bounded above and below on a ball about $x_0$.

Choose concentric balls $B(x_0,r)\subset B(x_0,2r)\subset U$ on which $m\le f\le M$. For distinct $x,y\in B(x_0,r\mathrm{/}2)$, extend the segment from $x$ through $y$ to $z=y+(r\mathrm{/}2)(y-x)\mathrm{/}\mathrm{\parallel }y-x\mathrm{\parallel }$, which lies in $B(x_0,2r)$. Put $\theta =\mathrm{\parallel }y-x\mathrm{\parallel }\mathrm{/}(r\mathrm{/}2+\mathrm{\parallel }y-x\mathrm{\parallel })$, so $y=(1-\theta )x+\theta z$. Convexity yields $f(y)-f(x)\le \theta (M-m)\le (2(M-m)\mathrm{/}r)\mathrm{\parallel }y-x\mathrm{\parallel }$. Repeating with $x,y$ interchanged gives the reverse bound. Negation proves the concave case.