He defined a strict contraction $f:X\to X$ on a metric space $(X,d)$ as requiring $d\left(f\right(x),f(y\left)\right)\le cd(x,y)$ for $c\in (0,1)$ and for all $x,y\in X$. Your proposed solution doesn’t fix such a $c$; in fact, as $x\to 0;x\in (0,1]$, $c\to 1$, which is why $f^{\prime }\left(0\right)=1$.

Claim: You can’t solve the exercise

Proof (thanks to TheMajor). Let $\{x_n{\}}_{n=1}^{\infty }$ be a sequence in the domain converging to $x\in [a,b]$ such that $f^{\prime }\left(x\right)={lim}_{n\to \infty }\frac{f\left(x_n\right)-f\left(x\right)}{x_n-x}$. Since $f$ is a strict contraction with contraction constant $c$, $\forall n\in N^{+}:\frac{|f\left(x_n\right)-f\left(x\right)|}{|x_n-x|}\le c<1$. Since the absolute value is continuous, we conclude that $|f^{\prime }\left(x\right)|\le c<1$. ◻️