Problem of continuous real valued function
I don't understand Jim's comment on (b). He took $g(x)=f(x+1)-f(x)$. Why should there is a $c\in [0,1]$: $f(c)=f(c+1)$.
I can't apply mean value theorem here directly. Since given function need not be differentiable. I am not able to find a condition of IVT too.
Regarding the answer to (c)I understood.
$g(0) = f(1)-f(0)$ while $g(1)=f(2)-f(1) = f(0)-f(1)=-g(0)$.
So if $g(0)$ is positive, $g(1)$ is negative and vice versa (or both are $0$ already).
The intermediate value theorem (IVT) implies that a continuous $g$ must have a point where it assumes $0$ inbetween those points.