I will try to show you why this is intuitive using counter example. Let us call the set reachable by unsaturated paths A, and the other set B. so s is in A, and t is in B. We know that by definition, each edge from A to B is in the minimum cut. Now suppose A were some other set such that it had a node v such that v was not reachable by an unsaturated path. This means that there is an edge such that the path from s to v is saturated, and let its capacity be x and its endpoints be x1 and x2. Also, let the minimum capacity of the edge from v to B be y. If we move x2 to B, and all vertices on x2-v path to B, then we will have a more minimal cut than earlier, since the capacity of our cut is now reduced by y and increased by x (y-x>=0). Therefore it couldnt have been a minimum cut in the first place. The other case is that some edge from A to B is not saturated. This implies that there is a node in B which is reachable from A by an augmenting path, which again leads to a contradiction.