Let the function $f:[0,1]\rightarrow \mathbb{R}$ be continuous on $[0,1]$ and differentiable on $(0,1)$ and $|f'(x)|<1$ for $ \forall x \in (0,1)$
I want to prove this statement: there exist at most one $c\in[0,1]$ such that $f(c)=c$.
Proof 1: Suppose on the contrary that there are two distinct points such that $f(c)=c$ and $f(d)=d$ then by MVT there exist a point $e$ such that $f'(e)=1$ which is a contradiction so we have at most one fixed point.
Also note that we can find a function like $f(x)=x+{1\over 2}$ which satisfy all the property but does not have any fixed point.
Ignore the example. It is incorrect. We can instead take constant function.
Your proof is incomplete, since you did not explain how to deduce from the mean value theorem that such a point $e$ must exist. It is simple, though:$$1=\frac{f(c)-f(d)}{c-d}=f'(e),$$for some $e$.
And your example is not an example since it is not a map from $[0,1]$ into itself.