Functions with bounded derivatives on a open convex set The following is a theorem from baby Rudin:

Suppose $f:E\to {\bf R}^m$ be a differentiable function, where $E\subset {\bf R}^n$ is convex and open, and there is $M>0$ such that
  $$
\|f'(x)\|<M
$$
  for every $x\in E$. Then 
  $$
|f(b)-f(a)|\leq M \ |b-a|
$$
  for all $a,b\in E$. 

Here is my question:

Can one replace "convex" with "connected" in this theorem?

When $n=1$, this is obviously still true. But when $n\geq 2$, the method in the proof, which uses convexity to reduce the dimension to $1$, would fail if one has connectedness instead of convexity. But I don't have a counterexample. 
 A: Take an annulus with a radius removed, say
$$A = \{ z : 1 < \lvert z \rvert < 2\} \cap (\mathbb{C}\setminus [0,+\infty)).$$
As a function, take for example the argument (with values in $(0,2\pi)$). That is differentiable, the derivative is bounded, but since the argument jumps at the positive real axis, and we can come arbitrarily close to it from above and from below, there is no $M < +\infty$ with
$$\lvert \arg z - \arg w\rvert \leqslant M\cdot \lvert z-w\rvert$$
for all $z,w\in A$.
Domains where the boundedness of the derivative implies Lipschitz continuity cannot be too far from convex.
A bound on the derivative, $\lVert f'(x)\rVert \leqslant M$ implies
$$\lvert f(x) - f(y)\rvert \leqslant M \cdot \inf \left\{L(\gamma) : \gamma \text{ is a piecewise differentiable path from } x \text{ to } y \text{ in } E\right\},$$
as follows from
$$f(y) - f(x) = \int_0^1 f'(\gamma(t))\cdot \gamma'(t)\,dt$$
if $\gamma \colon [0,1] \to E$ is such a path. If the quotient
$$\frac{\inf \left\{L(\gamma) : \gamma \text{ is a piecewise differentiable path from } x \text{ to } y \text{ in } E\right\}}{\lvert y-x\rvert}$$
is bounded for all $x,y\in E$, then the boundedness of the derivative implies Lipschitz continuity.
