Consider
$$x_{n+1} =f \left( x_{n} \right) = \begin{cases} x_{n} & 0\leq x_{n}< 0.5 \\ x_{n}-1 & 0.5\leq x_{n}\leq 1 \end{cases}.$$
Restricting our case to the interval $0\leq x_{n}\leq 0.5$:
For any ‘small’ $\epsilon$ infinitesimally close to $x$:
$$\left \| x-\left ( x+\epsilon \right ) \right \|=\left | \epsilon \right |$$
But why does
$$\left | f\left ( x \right )-f\left ( x+\epsilon \right ) \right |=\left | 2\epsilon \right |?$$
That factor $2$ is wrong.
For $x < 0.5$ and $x+ε<0.5$, we have:
$$ \left| f(x) - f(x+ε) \right| = \left | x- (x+ε) \right | = \left | ε \right |$$
More generally, for sufficiently small $ε$, you have:
$$ \frac{f(x+ε)-f(x)}{ε} = f'(x),$$
which you only need to rearrange to find $\left| f(x) - f(x+ε) \right|.$