Answer
See proof
Work Step by Step
Take:
$$h(x)=f(x)-x$$
$$h'(x)=f'(x)-1\ne 0$$
Using the mean value theorem, there $c$ in $(a,b)$ such that:
$$h'(c)=\frac{h(b)-h(a)}{b-a}\ne 0 \to h(b)-h(a)\ne0 \to f(b)-b-(f(a)-a)\ne0\to f(b)-b\ne f(a)-a $$
If we suppose that $f$ has two fixed points $a$ and $b$ thus:
$$f(b)-b\ne f(a)-a \to 0\ne0 $$ which is false so the function $f$ has at most one fixed point.
