A smooth function with zero differential is constant on each connected component

Let $C$ be a connected component of $M$ and fix $p\in C$. We claim that the fiber $S:=\{q\in C:f(q)=f(p)\}$ is open in $C$.

Let $q\in C$. Choose a smooth chart $(U,x)$ around $q$ and an open Euclidean ball $B$ with $x(q)\in B\subseteq x(U)$; put $W:=x^{-1}(B)$. For any $r\in W$, the map ${\gamma }_{q,r}(t):=x^{-1}((1-t)x(q)+tx(r))(0\le t\le 1)$ is a smooth curve in $W$ from $q$ to $r$ by [F1]. For each $t_0\in [0,1]$, apply [L1] to the shifted curve $s\mapsto {\gamma }_{q,r}(t_0+s)$ at $s=0$; since $d{f}_{{\gamma }_{q,r}(t_0)}=0$, this gives $(f\circ {\gamma }_{q,r}{)}^{\mathrm{\prime }}(t_0)=0$. Thus [L2] makes $f\circ {\gamma }_{q,r}$ constant on $[0,1]$, so $f(r)=f(q)$. Therefore $f$ is constant on $W$.

Step 1.2 shows that every fiber of $f{\mathrm{\mid }}_C$ is open in $C$. Hence $S$ is open, and so is its complement $C\setminus S$, which is the union of the other fibers. Since $C$ is connected and $S$ is nonempty, one must have $C\setminus S=\mathrm{\varnothing }$. Therefore $f$ is constant on $C$.

Because the connected component $C$ was arbitrary, $f$ is constant on each connected component of $M$.