Q:

prove a function is a contraction

Suppose that $f: [0,1] \rightarrow [0,1]$ is continuously differentiable.  Further, suppose that $f$ has a fixed point $x_{0} \in (0,1)$ such that $|f'(x_{0})| < 1$.  Then there exists an open interval $I$ containing $x_{0}$ such that $\{  f^{n}(x) \}_{n}$ converges to $x_{0}$ for all $x \in I$. 
I know that I need to prove that $f$ is a contraction using the fact that $|f'(x_{0})|< 1$, and then use the Contraction Mapping Principle in the following way.  But I am not sure about the details in proving that $f$ is a contraction.  Maybe something related to the Mean Value Theorem...?
Let $\epsilon > 0$, and $c = |f'(x_{0})|$.  By continuity, $\exists$ a $\delta > 0$ such that if $x \in (x_{0} - \delta, x_{0} + \delta) $ then 
$$ \lvert f(x) - f(x_{0}) \rvert < \epsilon(1-c)  $$
Let $I = (x_{0} - \delta, x_{0} + \delta)$.  By the Contraction Mapping Principle, for each $n$, and $x \in I$,
$$ \lvert f^{n}(x) - x_{0} \rvert \leq \dfrac{c^{n}}{1-c} \lvert f(x) - x_{0} \rvert = \dfrac{c^{n}}{1-c} \lvert f(x) - f(x_{0}) \rvert < \dfrac{c^{n}}{1-c}\epsilon (1-c) = c^{n} \epsilon < \epsilon$$ 
since $x_{0}$ is a fixed point and $c^{n} < 1$.

A:

Hint: 

Find a closed interval $I$ containing $x_0$ such that $|f'(x)|<1-\epsilon$ for all $x\in I$.
Show that $f:I\to I$ (using the fact that $x_0$ is a fixed point of $f$, and $|f'(x_0)|<1$, combined with the mean value theorem). 
Use the mean value theorem again to show that $f$ is a contraction on $I$: $$|f(x)-f(y)|\le(1-\epsilon)|x-y|,\quad \forall x,y\in I.$$

