$F(f)＝\{x∈[0,1]\mid f(x)=x\}$ is finite set Let $f(x)$ be real valued $C^1$ function and $0\leq x\leq 1$ implies $\leq f(x)\leq 1$ holds.
And for all $x∈F(f)$, $f'(x)\not=1$ holds.
Then, I want to prove $F(f)＝${$x∈[0,1]\mid f(x)=x$} is finite set .
I noticed $F(f)$ is closed subset of $[0,1]$ because equalizer of continuous function to Hausdorff space is closed, and from intermediate value theorem, $F(f)$ cannot be empty set. But I cannot proceed from here.
Thank you in advance.
 A: $F(f)＝\{x\in [0,1]｜f(x)＝x\}$ is a compact set (it is closed and bounded). Therefore if $F(f)$ is infinite, then it has a limit point $x_0\in F(f)$ and we can find a strictly monotone sequence $(x_n)_n$ in $F(f)$ such that $x_n\to x_0$.
Now we apply Lagrange theorem to each interval $[x_k,x_{k+1}]$: there exists $y_k\in (x_k,x_{k+1})$ such that
$$f'(y_k)=\frac{f(x_{k+1})-f(x_{k})}{x_{k+1}-x_k}=\frac{x_{k+1}-x_{k}}{x_{k+1}-x_k}=1.$$
It follows that $y_k\to x_0$ and by continuity of $f'$,  $f'(y_k)\to f'(x_0)=1$ which is a contradiction.
P.S. In order to show that $f'(x_0)=1$, we may also note that since $x_0\in F(f)$ then $f(x_0)=x_0$ and as $k\to \infty$,
$$1=\frac{x_{k}-x_{0}}{x_{k}-x_0}=\frac{f(x_k)-f(x_0)}{x_k-x_0}\to f'(x_0).$$
A: If $F(f)$ is an infinite  subset of $[0,1]$ then there exists $x_0\in \overline {F(f)\setminus \{x_0\}}.$ Then $f(x_0)=\lim_{x\to x_0; x_0\ne x\in F(f)}f(x)=\lim_{x\to x_0;\, x_0\ne x\in F(f)}x=x_0.$
So $x_0\in F(f).$
But $f'(x_0)=\lim_{x\to x_0;\, x_0\ne x\in F(f)}\frac {f(x)-f(x_0)}{x-x_0}=1.$
