Answer
Take a function $g(x)=f(x)-x$ and apply the Intermediate Value Theorem to show that there exists a value $c\in[0,1]$ that $g(c)=0$.
Work Step by Step
We examine a function : $$g(x)=f(x)-x$$
From the exercise, we know that $f(x)$ is continuous on $[0,1]$, and function $y=x$ is also continuous on $[0,1]$. According to the properties of continuous functions, $g(x)=f(x)-x$ is also continuous on $[0,1]$.
According to the exercise, on $[0,1]$, $0\le f(x)\le1$
Therefore, $f(0)-0\ge0$, while $f(1)-1\le0$.
That means, for $x=0$, $g(0)\ge0$, and for $x=1$, $g(1)\lt0$
According to the Intermediate Value Theorem, since $g(x)$ is continuous on $[0,1]$, there exists a value $x=c\in[0,1]$ such that $g(c)=0$, or in other words, $$f(c)-c=0$$ $$f(c)=c$$
