Solutions

Since $f$ is continuous on $[0,1]$ and $f(x)$ is in $[0,1]$ for each $x$, we know that $f(x_0)=x_0$ for some $x_0\in[0,1]$ by exercise 7-11. In this exercise, we also have that $f'(x)\ne 1$ for all $x\in[0,1]$.

We now prove that $f(y)=y$ for exactly one $y\in[0,1]$. For contradiction, suppose there exists $x_1\in[0,1]$ in addition to $x_0$ such that $f(x_1)=x_1$, just as $f(x_0)=x_0$. Without loss of generality, assume $x_0<x_1$. We thus have by the mean value theorem that

$$f'(c)=\frac{f(x_1)-f(x_0)}{x_1-x_0}=\frac{x_1-x_0}{x_1-x_0}=1$$

for some $c\in(x_0,x_1)$. But this contradicts the supposition that $f'(x)\ne 1$ for all $x\in[0,1]$.