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&cup;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&cup;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.