Proving or disproving basic facts about sequences in Real Analysis I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.
\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):

(a) sequences $(x_n)$ and $(y_n)$, which both diverge, but whose sum $(x_n + y_n)$ converges;
(b) sequences $(x_n)$ and $(y_n)$, where $(x_n)$ converges, $(y_n)$ diverges, and $(x_n + y_n)$ converges;
(c) a convergent sequence $(b_n)$ with $b_n \ne 0$ for all $n$ such that $1/b_n$ diverges;
(d) an unbounded sequence $(a_n)$ and a convergent sequence $(b_n)$ with $(a_n - b_n)$ bounded
(e) two sequences $(a_n)$ and $(b_n)$, where $(a_n b_n)$ and $a_n$ converge but $(b_n)$ does not.

Solution.
(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.
Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.
(b) This request is impossible. If the $(x_n + y_n)$ is to be convergent, it implies we are able to make the distance $\vert{(x_n + y_n) - (x + y)}\vert$ as small as like. However, we cannot make $y_n$ to lie eventually in a set $(y - \epsilon, y + \epsilon)$. Hence, the sum cannot be convergent.
(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.
(d) This request is impossible. The key here is to show that, assuming here $(a_n)$ is bounded leads to the contradiction, leading to their difference also being bounded.
If $(a_n)$ is a bounded sequence, there exists a large number $M > 0$, such that $\vert{a_n}\vert < M$ for all $n \in \mathbf{N}$. If $(b_n)$ is a bounded sequence, there exists a large number $N > 0$, such that $\vert{a_n}\vert < N$ for all $n \in \mathbf{N}$.
Thus,
\begin{align*}
    \vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\
    &\le \vert{a_n}\vert + \vert{-b_n}\vert\\
    &< M + N
\end{align*}
(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not.

$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I expanded my original attempt with short proofs, proving/disproving each of the statements. 
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.

(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.
Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.
Short proof.
Consider $a_n = \sqrt{n + 1} - \sqrt{n}$.
Observe that,
\begin{align*}
\sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\
&=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\
&< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}}
\end{align*}
Pick $\epsilon > 0$. We can choose $N > \frac{4}{\epsilon^2}$. To show that this choice of $N$ indeed works, we prove that, that for all $n \ge N$,
\begin{align*}
\absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\
&< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon
\end{align*}
Thus, $(\sqrt{n+1} - \sqrt{n}) \to 0$.
Consider $b_n = n - \sqrt{n^2 + 2n}$
Observe that:
\begin{align*}
n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\
&= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\
&< \frac{1}{n + \sqrt{n^2}} = \frac{1}{2n}
\end{align*}
Pick an arbitrary $\epsilon > 0$. We choose an $N > \frac{1}{2\epsilon}$. To show that choice of $N$ indeed works, we find that:
\begin{align*}
\absval{n - \sqrt{n^2 + 2n} - (-1)} &< \frac{1}{2n} \\
&< \frac{1}{2} \cdot (2\epsilon) = \epsilon
\end{align*}
Thus, $(n - \sqrt{n^2 + 2n}) \to -1$.
(b) This request is impossible. Alternatively, we believe that if $(x_n)$ converges and $(x_n + y_n)$ converges, then $(y_n)$ converges. Let us prove this fact rigorously. Suppose $\lim x_n = a$ and $ \lim x_n + y_n = b$. We shall prove that $\lim y_n = b - a$.
Observe that,
\begin{align*}
\absval{y_n - (b-a)} &= \absval{(x_n + y_n) - x_n - (b-a)}\\
&= \absval{(x_n + y_n - b) - (x_n - a)}\\
&\le \absval{x_n + y_n - b} + \absval{x_n - a}
\end{align*}
Pick an $\epsilon > 0$. Since $(x_n) \to a$, we can make the distance $\absval{x - a}$ as small as we like. There exists an $N_1$ such that
\begin{align*}
\absval{x_n - a} < \frac{\epsilon}{2}
\end{align*}
for all $n \ge N_1$.
Since $(x_n + y_n) \to b$, we can make the distance $\absval{x_n + y_n - b}$ as small as we like. There exists an $N_2$ such that,
\begin{align*}
\absval{x_n + y_n - b} < \frac{\epsilon}{2}
\end{align*}
Let $N = \max \{N_1,N_2 \}$. To show that this $N$ indeed works, we find that:
\begin{align*}
\absval{y_n - (b-a)} &\le \absval{x_n + y_n - b} + \absval{x_n -a}\\
&< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon
\end{align*}
(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.
Consider $(b_n) = \frac{1}{n}$. Pick an arbitrary $\epsilon > 0$. We can choose $N > \frac{1}{\epsilon}$. To show that this choice of $N$ indeed works, we find that:
\begin{align*}
\absval{\frac{1}{n}} < \epsilon
\end{align*}
for all $n \ge N$. Consequently, $(1/n) \to 0$.
(d) This request is impossible.
(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not. Let us prove $\frac{\sin n}{n}$ converges to $0$.
Observe that,
\begin{align*}
\absval{\frac{\sin n}{n}} &= \frac{\absval{\sin n} }{\absval n}\\
&\le \frac{1}{n}
\end{align*}
Let $\epsilon > 0$ be an arbitary small but fixed positive real number. We choose an $N > \frac{1}{\epsilon}$. To prove that this choice of $N$ indeed works, we have,
\begin{align*}
\absval{\frac{\sin n}{n}} &\le\frac{1}{n} \\
&< \frac{1}{(1/\epsilon)} = \epsilon
\end{align*}
Consequently, $\frac{\sin n}{n} \to 0$.
