The statement of the question is:
Suppose $f$ maps the open interval $E$ into itself, $0 < b < 1$, $f$ has property $X(b)$ (that property is Lipschitz continuous), and $x_0 \in E$ Prove that the sequence $\{ x_k\}$ defined recursively by $x_k = f(x_{k−1})$ for $k \geq 1$ converges.
What I have thus far :
Given $x_0 \in E$, we have $ x_n \in E$, $\forall n \in \mathbb{N}$
let $n = 0, 1, ..., k-1$
Then we can rewrite the given property $X(b)$ and $f(x_{k-1}) = x_k$ as:
$|x_{ k+1 } - x_k | = |f(x_k) - f(x_{ k - 1})| \leq b |x_k - x_{ k - 1}| $
And then by induction
$|x_{k+1} - x_k| \leq b |x_k - x_{ k-1}| \leq ... \leq b ^k |x_1 - x_0|$
Thus,
$ |x_{k+1} - x_k| \leq b^k |x_1 - x_0 | $
Note that given $x_0$, then $x_1 - x_0$ is a constant and $|b| < 1 $ then $b^k \rightarrow 0$ as $k \rightarrow \infty$ and thus the entire expression $ b^k | x_1 - x_0 | $ will tend to zero as $k \rightarrow \infty$ and so,
$$\lim_{k\to\infty}|x_{k+1}-x_{k}| \leq 0$$
I want prove what I have is Cauchy but I am not sure what to set my variables? or how to tie it all together...
The previous answer was incorrect. What that answer proved is: For every $\epsilon > 0$ and every $n$, there is some $N$ such that for every $k \ge N$, $|x_{k+n} - x_k| < \epsilon$. But to prove that the sequence is Cauchy, you need to know that for every $\epsilon > 0$ there is some $N$ such that for every $k \ge N$ and every $n$, $|x_{k+n} - x_k| < \epsilon$. In other words, $N$ cannot depend on $n$.
To fix the proof, note that \begin{align*} |x_{k+n} - x_k| &= |(x_{k+1} - x_k) + (x_{k+2} - x_{k+1}) + \cdots + (x_{k+n} - x_{k+n-1})|\\ &\le |x_{k+1} - x_k| + |x_{k+2} - x_{k+1}| + \cdots + |x_{k+n} - x_{k+n-1}|. \end{align*} Now fill in your bound for each absolute value of a difference and use the formula for the sum of a finite geometric series.