The purpose of this answer is to provide some supplementary insight into Chris Eagle's answer and to show in particular that his proof of MVT $\implies$ LUB is a very natural one.
In $\S 4$ of this note, I discuss the notions of completeness and Dedekind completeness in linearly ordered sets in terms of the induced order topology. (Recall an ordered set is Dedekind complete if it satisfies LUB -- every nonempty subset which is bounded above has a least upper bound -- and is complete if it is Dedekind complete and has a minimal element and a maximal element.) In particular, I show that a linear order is dense and Dedekind complete iff it is connected in the order topology. Every ordering on a field (compatible with the field structure, of course) is dense, so we see that an ordered field satisfies LUB iff it is connected in the order topology.
(In fact Chris Eagle's answer gives a snappy proof of "connected $\implies$ LUB": if LUB does not hold, take a nonempty set $A$ which is bounded above but has no least upper bound. Then the set of upper bounds of $A$ is nonempty, proper, open and closed. The other implication is the one which is more familiar from real analysis, and is not quite so easy. In the above note, I take the perspective that a very natural proof is given using the technique of real induction.)
So if we want to show that MVT $\implies$ LUB (as Chris Eagle points out, LUB $\implies$ Rolle's Theorem $\implies$ MVT is part of the standard honors calculus / elementary real analysis curriculum), it is natural to go by contrapositive: if an ordered field $(F,+,\cdot,<)$ does not satisfy LUB, then it is not connected, i.e., there exists a nonconstant continuous function $f: F \rightarrow \{0,1\}$. It follows immediately from the definition of the order topology that for each $x \in F$, there exists an open interval $I$ containing $x$ such that $f|_I$ is constant. Thus $f$ is a nonconstant function which is everywhere differentiable with derivative identically zero, contradicting MVT.