Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.
Let $\{x_n\}$ be recursively defined as $ \left\{ \begin{array}{ll} x_{n+1} =\frac{1-x_n}{4} \\ x_1 = \frac{1}{2} \end{array} \right. $
$(a)$ Prove that $|x_{n+1}-x_n| \le \frac{1}{4}|x_n-x_{n-1}|$
$(b)$ Use $(a)$ to prove that ${x_n}$ is a cauchy sequence
$(c)$ Compute $\underset{n \rightarrow \infty}{\rm lim} x_n$
Answer:
$(a)$ $x_{n+1} - x_n=\frac{1-x_n}{4}-x_n$ and $ x_n = \frac{1-x_{n-1}}{4}$. Thus, $x_{n+1}-x_n = (\frac{1-x_n}{4})-(\frac{1-x_{n-1}}{4}) = \frac{1}{4} \{1-x_{n}-1+x_{n-1}\}$. Therefore, $|x_{n+1} -x_n| \le |\frac{1}{4}(1-x_n-1+x_{n-1})| \Rightarrow |x_{n+1} -x_n| \le \frac{1}{4} |x_{n-1}-x_n|$ and $\frac{1}{4} |x_{n-1}-x_n| = \frac{1}{4} |x_{n}-x_{n-1}|$
$(b)$ From $(a)$, it is observable that $|x_{n+1}-x_n| \le \frac{1}{4}|x_{n-1}-x_n|$. Now $|x_{n-1}-x_{n-2}| \le \frac{1}{4} |x_{n-2} - x_{n-3}|$. Continuing this way, we get $|x_{n+1}-x_n| \le \frac{1}{4^n} |x_2-x_1| \Rightarrow |x_{n+1}-x_n| \le \frac{1}{4^n}(\frac{+3}{8})$. As $x \rightarrow \infty , \frac{1}{4^n} \rightarrow 0$. Hence, $\{x_n\}$ becomes a Cauchy sequence. Since every sequence in $\mathbb{R}$ is convergent, $\{x_n\}$ is convergent.
$(c)$ In $(b)$ we have shown $\{x_n\}$ is convergent. Therefore, we assume $l \in \mathbb{R}$ and $ lim\hspace{3pt} x_n = l$. We know $x_{n+1} = \frac{1}{4} (1-x_n)$; $lim \hspace{3pt} x_{n+1} = \frac{1}{4}(1-lim\hspace{3pt} x_n) \Rightarrow l=(\frac{1}{4})(1-l) \Rightarrow l=\frac{1}{5}$
For (a) and (c) you proofs are correct But your argument for (b) is wrong. $x_{n+1}-x_n \to 0$ does not imply that $(x_n)$ is a Cauchy sequence. Use the fact that $|x_{n+m}-x_n| \leq |x_{n+m}-x_{n+m-1}|+|x_{n+m-1}-x_{n+m-2}|+\cdots +|x_{n+1}-x_{n}|$ to show that $x_{n+m}-x_n \to 0$ as $n,m \to \infty$. [You will have to compute a geometric sum for this].