Let $M$ and $N$ two manifolds which have the same dimension, $f:M\to N$ a map $\mathcal{C^\infty}$. We suppose that $M$ is compact and we have $b$ a regular value of $f$.
First, I have to prove that $f^{-1}(\{b\})$ is a compact and finite set.
For compacity, I thought about the topology of $\mathbb{R}^m$ using the fact that if a set is closed and bounded it's compact. I tried to define a map $\mathbb{R}^m\to \mathbb{R}^m$ using homeomorphism $\phi$ and $\psi$ but I don't know if it is a good method. To prove that it is finite I have no idea.
Second, I have to prove that there exists an open set $V$ of $N$ which contains $b$ such as for all $c\in V$ we have $\mid f^{-1}(\{c\}) \mid\ge\mid f^{-1}(\{b\}) \mid$.
Thanks in advance !
Since $f$ is continuous, $f^{-1}(b)$ is closed. It is compact since it is a closed subset of a compact set.
Let $x\in f^{-1}(b)$, since $f$ is regular, $df_x:T_xM\rightarrow T_bN$ is an isomorphism. You can work locally and restrict $f$ to a neighborhood $U$ of $x$ such that $f(U)\subset V$ where $V$ is a neighborhood of $b$ and $U$ and $V$ are diffeomorphic to open subsets of $R^n$. This implies that the differential $f_{\mid U}:U\rightarrow V$ is invertible so it is a local diffeomorphism. We deduce that there exists an open subset $x\in U_x$ such that the restriction of $f$ to $U_x$ is a diffeomorphic onto its image. This implies that $f^{-1}(b)\cap U_x=\{x\}$. Thus $f^{-1}(b)$ is discrete. A discrete compact set is finite.