$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I'm self-learning mathematical analysis from Understanding Analysis by Stephen Abbott. I'd like someone to verify my solution, if the proof checks out.
Additionally, if $(a_n)$ is Cauchy, is $[[a_n]]$ also Cauchy? Because, there is no strict relationship between $\absval{[[a_n]] - [[a_m]]}$ and $\absval{a_n - a_m}$ or so I think.
Exercise 2.6.4
Let $(a_n)$ and $(b_n)$ be Cauchy sequences. Decide whether each of the following sequences is a Cauchy sequence, justifying each conclusion.
(a) $c_n = \absval{a_n - b_n}$
(b) $c_n = (-1)^n a_n$
(c) $c_n = [[a_n]]$, where $[[x]]$ refers to the greatest integer less than or equal to $x$.
Proof.
(a) The following inequalities hold for any reals $a,b \in \mathbf{R}$: \begin{align*} \absval{a + b} \le \absval{a} + \absval{b} \end{align*} Replacing $b$ by $-b$, we have: \begin{align*} \absval{a - b} \le \absval{a} + \absval{b} \end{align*} Writing $a = a - b + b$, we have: \begin{align*} \absval{a} &= \absval{a - b + b}\\ &\le \absval{a - b} + \absval{b}\\ \iff \absval{a} - \absval{b} &\le \absval{a - b} \end{align*} Writing $b = b - a + a$, we have: \begin{align*} \absval{b} &= \absval{b - a + a}\\ &\le \absval{a - b} + \absval{a}\\ \iff -(\absval{a} - \absval{b}) &\le \absval{a - b} \end{align*} Consequently, \begin{align*} \absval{\absval{a} - \absval{b}} \le \absval{a - b} \end{align*}
Now, the distance between any two terms of the sequence $(c_n)$ is given by,
\begin{align*} \absval{c_n - c_m} &= \absval{\absval{a_n - b_n} - \absval{a_m - b_m}}\\ &\le \absval{{a_n - b_n} - ({a_m - b_m})} = \absval{(a_n - a_m) - (b_n - b_m)}\\ &\le \absval{a_n - a_m} + \absval{a_m - b_m} \end{align*}
We desire each of these distances to be smaller than $\epsilon/2$.
Since $(a_n)$ and $(b_n)$ are Cauchy sequences, given any $\epsilon > 0$, there exists $N_1,N_2$ such that for all $m,n \ge N_1$, $\absval{a_n - a_m} < \epsilon/2$ and for all $m,n \ge N_2$, $\absval{b_n - b_m} < \epsilon/2$. Altogether, define $N = \max \{ N_1,N_2 \}$, if $m,n \ge N$, the terms in the above inequality are each smaller than $\epsilon/2$, which is what we required.
(b) $c_n = (-1)^n a_n$
If $(a_n)$ is Cauchy, $((-1)^n a_n)$ is also Cauchy, because \begin{align*} \absval{(-1)^n(a_n - a_m)} = \absval{a_n - a_m} \end{align*}
(a) It is correct, but you can use the inequality $\bigl||a|-|b|\bigr|\leqslant|a-b|$ without proving it. It's a standard inequality. Besides, when you wrote $|a_n-a_m|+|a_m-b_m|$, what you meant was $|a_n-a_m|+|b_n-b_m|$.
(b) That is wrong. The constant sequence $1$ is a Cauchy sequence, but $\bigl((-1)^n\bigr)_{n\in\Bbb N}$ isn't.
(c) The statement is false. For instance, the sequence $\left(\frac{(-1)^n}n\right)_{n\in\Bbb N}$ is a Cauchy sequence, but$$(\forall n\in\Bbb N):\left\lfloor\frac{(-1)^n}n\right\rfloor=\begin{cases}-1&\text{ if $n$ is odd}\\0&\text{ if $n$ is even.}\end{cases}$$and therefore $\left(\left\lfloor\frac{(-1)^n}n\right\rfloor\right)_{n\in\Bbb N}$ is not a Cauchy sequence.