Solutions

For contradiction, suppose that $f(x)\ne x$ for all $x$. Define $g(x)=x$. Notice that $g(0)=0<f(0)$ because $f(0)\ne0$. Likewise, $g(1)=1>f(1)$ because $f(1)\ne1$. Hence, by Exercise 10, $g(x)=x=f(x)$ for some $x$ in $[0,1]$, which is a contradiction.