TITLE: Does ZFC prove the universe is linearly orderable?
QUESTION [16 upvotes]: It is consistent with ZFC that the universe is well-ordered, e.g. in $V=L$ where global choice holds. I also know that it is consistent that global choice fails (although I have no immediate example from the top of my head).
However one can try and ask a slightly weaker question, much like the axiom of choice implies every set can be linearly ordered, but the reversed implication fails; is it the same with global choice? The immediate answer is yes, it is consistent that the universe can be linearly ordered but the axiom of choice fails and therefore global choice has to fail too. But just how far can this be pushed?

Can we construct a model of ZFC in which the axiom of choice holds, but there is no linear ordering of the universe? Or does ZFC prove that the universe is linearly ordered?

One would expect that it would, because if we assume AC then the sets of ordinals already decide the universe in its entire form, and sets of ordinals can always be linearly ordered by: $$A\prec B\iff\min (A\triangle B)\in A.$$ (where $\triangle$ denotes the symmetric difference.)

REPLY [8 votes]: Joel nicely answers Asaf's question.
  Here I just want to add some footnotes
  to his answer.

1. Joel' argument shows that $ZFC$ cannot even prove the definable class form of axiom of choice for pairs. Historically, this was first done by with a similar argument by Easton in his 1964-thesis (printed in the Annals of Mathematical Logic).  
2. In the pre-forcing era, Mostowski had already shown that $ZFA$ + "the universe can be linearly ordered" does not imply "every set can be well-ordered" (see p.51 of Jech's book on the Axiom of Choice). 
3. As shown in the proof of Theorem 5.21 (p.71) of Jech's text, Mostwoski's argument can be transplanted to the forcing context to show that indeed $V$ can be definably linearly ordered in Cohen's so-called basic model of the failure of the axiom of choice.