Answer
See explanations.
Work Step by Step
Step 1. As point $(c, f(c))$ lies on the line $y=x$, we will need to show that the function $f(x)$ intersects with $y=x$ at least once.
Step 2. Let $g(x)=f(x)-x$. We need to show that the function $g(x)$ has at least one zero.
Step 3. Given $0\leq f(x)\leq 1$, there must exist two points $x_1, x_2$ in the interval $[0,1]$ such that $f(x_1)=0, f(x_2)=1$ which gives $g(x_1)=0-x_1=-x_1\lt0, g(x_2)=1-x_2\gt0$ (Please note that if $x_1=0$ or $x_2=1$, we automatically prove the case; thus, we assume $x_1\ne0$ and $x_2\ne1$ here.)
Step 4. Since $f(x)$ is a continuous function, $g(x)$ is also continuous. With $g(x_1)\lt0, g(x_2)\gt0$, by using the Intermediate Value Theorem, we know that there must be a point $c$ between $x_1$ and $x_2$ such that $g(c)=0$, which means that $f(c)=c$.
