I thought ZFC proved the existence of an inductive well-ordering that is itself a set for any stage of V. NBG with only the regular AC should then prove/assert the existence of a class R of ordered pairs **(a,b)** such that either: 


a has lesser rank than b, or:


they both have rank α and **(a,b)** exists in the ZFC well-ordering of the complement of V(α) within V(α+1). 


I don't see that we have to quantify over classes there. By another phrasing it just unites for all V(β) their inductive well-orderings along these lines, which are already supersets of the same ordering for any previous stage. And then NBG should easily prove this well-orders V. How does this proof fail?



Why doesn't choice imply global choice (in NBG)?
Your first sentence is true (modulo the word "inductive"), but not in the way you mean: $ZFC$ proves the existence of *many* set well-orderings of each $V\_\alpha$. Now, under some further assumption - say, $V=L$ - there might be a sequence of somehow canonical well-orderings of the $V\_\alpha$, in which case we can indeed "glue them together" to get the global well-ordering you seek. However, if there is no such definable sequence, then we can't define your well-ordering, since we would need to pick some specific well-ordering of each of the $V\_\alpha$s simultaneously - that is, find a choice function for a class-sized sequence of sets. And this, of course, is *exactly* global choice.


EDIT: One positive point. Even though choice doesn't imply global choice, $NBG$ is strongly conservative over $ZFC$, so in some sense $ZFC$ is not enough to show that a well-ordering of $V$ exists, but is enough to show that a well-ordering of $V$ isn't too destructive to the universe. :)


