= Solution
Choose $M$ with $|f|\leq M$ on $U$, and fix $r=(1-\|x\|)/4>0$. We prove a local <Lipschitz continuity> bound, which gives more than <continuity>. Let $u,v\in B(x,r)$, put $d=\|v-u\|$, and suppose $d>0$. The point $w=u+(2r/d)(v-u)$ lies in $B(x,3r)\subset U$, while $d<2r$ and
$$
v=\left(1-\frac d{2r}\right)u+\frac d{2r}w.
$$
By <convexity>, $f(v)-f(u)\leq[d/(2r)](f(w)-f(u))\leq Md/r$. Interchanging $u,v$ gives
$$
\boxed{|f(v)-f(u)|\leq\frac Mr\|v-u\|\qquad(u,v\in B(x,r)).}
$$
The case $d=0$ is immediate. Every point has such a neighborhood, so \b[$f$ is locally Lipschitz and therefore continuous throughout $U$]. This is the <bounded convex functions are locally Lipschitz on an open ball> argument; no finite-dimensional assumption is needed.
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