I've got a sequence $x_n$ such that I've proved $b\leq x_n \leq c$, and $|x_{n+1}-x_{n}|\leq \frac{4}{9}|x_n-x_{n-1}|$
However I'm not very familiar with Cauchy sequences, so I don't know how to exactly prove it's Cauchy. I've gotten so far:
Let $m,n>0$. Then $m=n+a$. So \begin{align*} |x_m-x_n| &= |x_{n+a}-x_n|\\ &= |(x_{n+a}-x_{n+a-1})+(x_{n+a-1}-x_{n+a-2})+\ ...\ +(x_{n+1}-x_n)|\\ &\leq a|x_n-x_{n-1}| \end{align*}
I know I'm close, but I'm not sure where to go from here.
EDIT: \begin{align*} |x_m-x_n| &= |x_{n+a}-x_n|\\ &= |(x_{n+a}-x_{n+a-1})+(x_{n+a-1}-x_{n+a-2})+\ ...\ +(x_{n+1}-x_n)|\\ &\leq \Sigma_{i=1}^{i=a}(\frac{4}{9})^i * |x_{n}-x_{n-1}|\\ &\leq (c-b)\frac{4}{9}\cdot \frac{1-(4/9)^a}{5/9}\\ &\leq \frac{4}{5}(c-b)(1-(\frac{4}{9})^a) \end{align*}
But now I'm not sure how to say it's Cauchy.
EDIT 2:
$$\frac{4}{5}(c-b)(1-(\frac{4}{9})^a) \leq \frac{4}{5}(c-b)$$
So let $n_0 = \epsilon * \frac{5}{4(c-b)}$. Then for any $\epsilon > 0$, $|x_m-x_n|<\epsilon$ whenever $m,n > \frac{5}{2}\epsilon$.
What you have is that $x_{n}$ is a contractive sequence, such that $|x_{n+1} - x_{n} | \le C |x_{n} - x_{n-1}|$ for all $n \ge N$, $0 < C < 1$. You have $C = \frac{4}{9}$. Theorem : A contractive sequence is a Cauchy sequence, so it is enough to use the theorem.
But if you also want to prove it (you should try), in general it is the same step as you have so far, then use the hint from @julien with geometric series of ratio $C$.