Perhaps I've misunderstood the question, but it looks like it's false.

Let M={(x,y)∈ℝ²|(x,y)≠(0,0)}, N=ℝ, and define f(x,y)=x. This is a smooth map of smooth manifolds, with the fibers over ℝ-{0} connected, but the fiber over 0 disconnected.

**Edit:** Wayne has added the hypothesis that M is compact. I think the statement is true under this hypothesis. Here's a sketch proof. Suppose f^{-1}(x) is disconnected, then I'd like to prove that there is an open neighborhood of x where the fibers are disconnected. Since manifolds are normal, there are two non-empty disjoint open sets U and V in M covering f^{-1}(x). Now prove a generalization of the hotdog lemma, which will say that there is an open neighborhood W of x such that U∪V covers f^{-1}(W). Since U and V are disjoint, this will show that the fibers over points of W are disconnected. To prove the generalized hotdog lemma. use the fact that smooth maps locally "look like products", choose a cover of f^{-1}(x) by "box shaped" open sets contained in U∪V. You can choose a finite number of these by compactness of f^{-1}(x) (it's a closed subset of a compact space), and take W to be the intersection of all of their images in N.

**More Edit:** The above proof doesn't work (see comments below and Richard Kent's post). Apparently, I'm confused about the meaning of Ehresmann's_theorem, because it looks to me like the map f:S^{2}⊂ℝ^{3}→ℝ given by f(x,y,z)=z is smooth, but it doesn't look like a trivial fibration around the poles. The algebro-geometric analogue says that a smooth morphism X→Y always factors as X→**A**^{n}_{Y}→Y, where the first map is etale. But an algebraic geometer would say that the map S^{2}→ℝ is not smooth.