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<sup>-1</sup>(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<sup>-1</sup>(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<sup>-1</sup>(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<sup>-1</sup>(x) by "box shaped" open sets contained in U∪V. You can choose a finite number of these by compactness of f<sup>-1</sup>(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][1]). Apparently, I'm confused about the meaning of [Ehresmann's_theorem][2], because it looks to me like the map f:S<sup>2</sup>⊂ℝ<sup>3</sup>→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→<b>A</b><sup><sup>n</sup></sup><sub>Y</sub>→Y, where the first map is etale. But an algebraic geometer would say that the map S<sup>2</sup>→ℝ is not smooth.
[1]: http://mathoverflow.net/questions/4054/can-connectedness-of-fibers-of-a-smooth-map-be-checked-on-a-dense-set/4189#4189
[2]: http://en.wikipedia.org/wiki/Ehresmann%27s_theorem