Let $c_n=\frac{1}{2}(a_n+b_n), r=\lim_{n\to \infty}c_n,$ and $e_n=r-c_n$. Here $[a_n,b_n]$, with $n\geq 0$, denotes the succesive intervals the arise in the bisection method when it is applied to a continuous function $f$.
(a) Show that $|e_n|\leq 2^{-n}(b_1-a_1)$
(b) Show that $e_n=O(2^{-n})$ as $n\to \infty$
(c) Is it true that $|e_0|\geq |e_1|\geq ...$? Explain.
(d) Show that $|c_n-c_{n+1}|=2^{-n-2}(b_0-a_0)$
For (a) $|e_n|=|r-c_n|\leq \frac{1}{2}|b_n-a_n|=\frac{1}{2}2^{-(n-1)}(b_1-a_1)=2^{-n}(b_1-a_1)$ because $b_{n+1}-a_{n+1}=\frac{1}{2}(b_n-a_n)$
For (b) We know that $|e_n|\leq 2^{-n}(b_1-a_1)$ for $n\geq 0$ and so, taking $C:=(b_1-a_1)>0$ we get that $|e_n|\leq C2^{-n}$ for all $n\geq 0$, which means that $e_n=O(2^{-n})$ as $n\to \infty$.
For (c) yes it is true because $(b_1-a_1)\geq (b_1-a_1)/2\geq (b_1-a_1)/4\geq...$.
For (d) Show that $|c_n-c_{n+1}|=2^{-n-2}(b_0-a_0)$.
Is this reasoning correct? Thank you very much.