A theorem in Morse theory If $M$ is a smooth manifold on which there is a smooth function $f:M \to ( - 1,2)$  such that all $[0,1]$ are regular values of $f$ and ${f^{ - 1}}(s)$ is a compact set for all $s \in [0,1]$，then is ${f^{ - 1}}([0,1])$ a compact set in $M$?
As we know, if we assume ${f^{ - 1}}([0,1])$ to be a compact set in $M$ as a condition, then according to a theorem in Morse theory, ${f^{ - 1}}(0)$ and ${f^{ - 1}}(1)$ are diffeomorphic. This proof consists of constructing a vector field and using the integral curves related to that field. The crucial part is that since ${f^{ - 1}}([0,1])$ is compact, we can always construct a vector with compact support, therefore the integral curves are complete. In this way the certain diffeomorphism can be defined without problem.
Amazingly enough, with the same conditions in the first paragraph, ${f^{ - 1}}(0)$ and ${f^{ - 1}}(1)$ seem to be still diffeomorphic. However, the same method, that is (constructing vector field), cannot be applied to this situation easily since the domain of integral curves are determined by ${M_t} = {f^{ - 1}}(t){\kern 1pt} {\kern 1pt} (t \in [0,1])$, not necessarily uniform. The deficiency can be overcome by proving ${f^{ - 1}}([0,1])$ is a compact set, this is where I got stuck, or constructing the integral curves more intricately
 A: I don't think the statement you are asking about in the beginning of your posting is true. Suppose $(M_0, g)$ is a compact smooth Riemannian manifold -- a one dimensional circle will do. Define $$ M = M_0 \times (-1,2) \cup M_0 \times (-1,1/2)$$ (disjoint union of cartesian products) with the metric $g_M$ defined trivially on the first factor: 
$$g_M((v,t),(w,s)) = g(v,w) + st$$ 
and as $$g_M((v,t),(w,s))=\lambda(p,z) g(v,w) + st$$ 
on the second factor. Here, $(p,z) \in M_0 \times (-1,1/2)$, $v, w \in T_p M_0$ and $s,t \in T_z (-1, 1/2)$ Choose $\lambda$ such that the diameter of $(M_0\times (z))$ tends to $ \infty$ as $z\rightarrow 1/2$. For the 2nd factor, think of something like the graph of $1/(1-|(x,y)|)$ outside of a sufficiently large ball. 
Now define $f$ simply as the height function, $f(p, z) = z$.
$M$ is smooth, $f$ is smooth, $f^{-1}(s)$ is compact for every $s\in [0, 1]$ and every point in the image is a regular value. $f^{-1} ([0,1]) $ is probably even a complete Riemannian manifold.
This example will clearly become invalid if you require connectedness. Is your $M$ connected?
Edit: connectedness of $M$ does not invalidate the counterexample, the two components from the example above may be smoothly glued together below the $0$ level. An extension of $f$ will then have critical points below the $0$ level, though. This is not excluded by the assumption from the original question, so this remains a valid counterexample.
Note: for this example, $f^{-1}(0)$ and $f^{-1}(1)$ are obviously not diffeomorphic. 
