Exercise 4.5.7

Since 0 and 1 are both in the range of $f$, choose $a$ and $b$ such that $f(a)=0$ and $f(b)=1$. Define $g(x)=f(x)-x$; clearly $g$ is continuous, $g(a)=0-a=-a\le 0$, and $g(b)=1-b\ge 0$. By IVT there must be some $c\in [a,b]$ so that $g(c)=0$ and hence $f(c)=c$.