Question. Let $\Omega$ be an open subset of $\mathbb R^3$, and let $f: \Omega \to \mathbb R$ be continuous. Then $f$ is
(a) Lipschitz, if it is $C^1$ and $\nabla f$ is bounded;
(b) uniformly continuous, if $\Omega$ is a disk;
(c) Lipschitz, if $\Omega = \mathbb R^3$;
(d) uniformly continuous, if it is differentiable, $\nabla f$ is bounded and $\Omega$ is a half-plane.
I know (a) is not justified, since we need further information about $\Omega$ (needs to be convex). Also the function $f: \mathbb R^3 \to \mathbb R$ such that $f(x,y,z) = x^n + y^n + z^n$ with $n \geq 2$ in the naturals seems to work as a counterexample to (c).
The main points I don't understand, though, are the requirements on $\Omega$ we find in (b) and (c): if $\Omega \subseteq \mathbb R^3$, how can it be open if it is a disk or a half-plane – in general, a "flat", 2D subset? Wouldn't all points in $\Omega$ be boundary points then?
At any rate, a disk without its boundary circumference (that's what I think they could be meaning when they say "open") is not compact, so Heine's theorem does not apply and (c) is not necessarily true, so I'm thinking (d) by exclusion, although I don't know which theorems would directly justify picking it as an answer. Of course, proving it is Lipschitz in those conditions would automatically prove it's uniformly continuous. Any input is appreciated!