Why is in above theorem, it is assumed that $D$ an open disk? Is it to make sure that we can surely apply mean value Theorem, ie, walking parallel to x or y axis, we are not moving out of domain? Can a convex open set do that job?
Also, this may sound silly, but why open disk? Is it so that derivative is defined without fuss, that is, without considering boundary of that disk??
The domain can in fact be an arbitrary open connected subset of $\mathbb C$ and the statement will still hold. Of course, the proof will have to be changed, because then we will not be able to go from an arbitrary point to another arbitrary point of the domain applying twice the mean value theorem (in other words: your guess was good).
The domain doesn't have to be open, but in order to have a sensible notion of derivative, at least the domain cannot have isolated points.