(1) Your proof that $\liminf A_n\subseteq \limsup A_n$ is correct.
Exercise 1: Under what conditions does equality hold in the above inclusion, i.e., under what conditions is it true that $\liminf A_n=\limsup A_n$?
(2) Note that $x\in \liminf A_n$ if and only if there exists a positive integer $N$ such that $x\in A_n$ for all $n\geq N$ if and only if there exists a positive integer $N$ such that $x\in \bigcap_{k=N}^{\infty} A_k$ if and only if $x\in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$.
I will leave the remaining questions as easy exercises (with similar solutions).
Exercise 2: Let $\{A_n\}$ be a sequence of measurable subsets of a measurable space $(X,\mu)$. Assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. Let $A=\limsup A_n$. Prove that $\mu(A)=0$. (Hint: Use the fourth assertion in your question, namely, use the characterization of $\limsup A_n$ in your question.)
Exercise 3: Give an example (in the context of Exercise 2) where $\lim_{n\to\infty} \mu(A_n)=0$ but that $\mu(A)>0$. Do not assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. (Hint: let $\{A_n\}$ be an appropriate sequence of intervals in $[0,1]$, for example.)
You ask whether or not your reasoning is correct. It is close, but not quite there.
Let's start from the beginning. We have $a_n = (-1)^n + 1/n$. Now, we can define (as you did), the sequence of sequences $$A_n = \{a_k\colon k \geq n\} = \{a_n, a_{n+1}, \ldots\}.$$
Note that each $A_n$ is not a number, but rather a sequence of numbers. (This makes your guess for $A_n$ incorrect.)
Now, by definition
$$b = \limsup a_n = \inf_{n\geq 1}\ \sup_{k\geq n} a_k = \inf_{n\geq 1}\sup\{a_k\colon k \geq n\} = \inf_{n\geq 1}(\sup A_n) = \inf_{n\geq 1}\, b_n$$
$$c = \liminf a_n = \sup_{n\geq 1}\ \inf_{k\geq n} a_k = \sup_{n\geq 1}\inf\{a_k\colon k \geq n\} = \sup_{n\geq 1}(\inf A_n) = \sup_{n\geq 1} \,c_n,$$
where we let $b_n = \sup A_n$ and $c_n = \inf A_n$, as you did.
So let's compute $b_n = \sup A_n$ and $c_n = \inf A_n$.
Claim: We claim that $$b_n = \sup A_n = \begin{cases} 1 + \frac{1}{n} \ \ \ \text{ if } n \text{ is even} \\ 1 + \frac{1}{n+1} \ \ \ \text{ if } n \text{ is odd} \end{cases}$$
and
$$c_n = \inf A_n = -1 \ \ \text{ for all } n \geq 1.$$
Proof: Let's look at $b_n$ first. If $n$ is even, then $$A_n = \left\{1 + \frac{1}{n}, \ -1 + \frac{1}{n+1}, \ 1 +\frac{1}{n+2}, \ldots\right\}.$$
It is clear that $A_n$ contains a largest element, namely $1 + \frac{1}{n}$. (If you don't find this fully precise, I invite you to fill in the details.) Therefore, $\sup A_n = 1 + \frac{1}{n}$ when $n$ is even.
If $n$ is odd, then $$A_n = \left\{-1 + \frac{1}{n}, \ 1 +\frac{1}{n+1}, \ -1 + \frac{1}{n+2}, \ldots\right\}.$$
Again, $A_n$ contains a largest element, namely $1 + \frac{1}{n+1}$, so that $\sup A_n = 1 + \frac{1}{n+1}$. This verifies the claim about $b_n$.
Let's look at $c_n$ now. We want to show that $\inf A_n = -1$, i.e. that $-1$ is the greatest lower bound of $A_n = \{a_k\colon k\geq n\}$.
Now, since $-1 < a_k$ for every $k \geq n$ (you can check this), it follows that $-1$ is a lower bound of $A_n$. To show that it is the greatest lower bound, we need to show that if $q > -1$, then $q$ is not a lower bound for $A_n$.
Let $q > -1$. Choose $k \geq n$ large enough so that $k > \frac{1}{q+1}$. Then $k(q+1) > 1$, so $q+1 > \frac{1}{k}$, so $q > -1 + \frac{1}{k}$. But $-1 + \frac{1}{k}$ is an element of $A_n$ (since $k \geq n$), and we just showed that it is less than $q$. Therefore, $q$ is not a lower bound for $A_n$. We therefore conclude that $\inf A_n = -1$ (for all $n \geq 1$).
This proves the claim.
Finally, it follows from our claim (I leave the details to you) that $$b = \limsup a_n = \inf_{n\geq 1}\ b_n = 1$$ and $$c = \liminf a_n = \sup_{n\geq 1} \ c_n = -1.$$
Best Answer
As soon as you write "Then by the definition of infimum", you are saying incorrect things.
First: if $b$ is the infimum of $A$, then for every $\epsilon\gt 0$ there exists at least one $a\in A$ such that $a\lt b+\epsilon$. However, you claim this holds for all elements of $A$, which is false. Take $A = (0,1)$. Then $\inf A = 0$; and it is indeed true that for every $\epsilon\gt0$ there exists at least one $a\in A$ with $a\lt\epsilon$; but if $\epsilon=\frac{1}{10}$, it is certainly false that every $a\in A$ is smaller than $\epsilon$.
Second: the elements of $A$ are not $a_n$s! They are suprema of infinite sequences of $a_n$s, and as such cannot be assumed to be $a_n$s. For example, if $a_n = 1-\frac{1}{n}$, then $A=\{1\}$, and no $a_n$ is equal to any element of $A$.
So that sentence is not just wrong, it's doubly wrong. The rest of course is now nonsense.
The second part does not seem to be proving anything; you are just asserting things. Why do the conditions imply the inequalities? What properties of the limit superior are you using? It's a mystery.
Rather: let $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$. Prove that $A_n$ is a decreasing sequence: $A_{n+1}\leq A_n$ for all $n\in \mathbb{N}$. Your set $A$ is precisely the set of $A_n$s.
Now let $b= \inf A = \inf\{ A_n\}$. By the definition of infimum, for every $\epsilon\gt 0$ there exists $N$ such that $b\leq A_N\lt b+\epsilon$. Since the sequence of $A_n$s is decreasing, then for all $n\geq N$ we have $b\leq A_n\leq A_N\lt b+\epsilon$, so in fact we have that $A_n\lt b+\epsilon$ ultimately. Moreover, since $a_n\leq A_m$ for all $n\geq m$, this implies that $a_n\lt b+\epsilon$ ultimately, as required.
For the second clause of the first part, let $\epsilon\gt 0$. Then $b-\epsilon\lt A_n$ for all $n$. Now remember what $A_n$ is. $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$; since $b-\epsilon\lt A_n$, there exists $k\geq n$ such that $b-\epsilon\lt a_k\leq A_n$. That is: for all $n$, there exists $k\geq n$ such that $a_k\gt b-\epsilon$, so $a_k\gt b-\epsilon$ frequently.
For part (2), let $b$ be a real number that satisfies the given properties. Since for every $\epsilon\gt 0$ we have that $b-\epsilon \lt a_n$ frequently, that means that $b-\epsilon$ is not an upper bound for $\{a_k\mid k\geq n\}$ for any $n$. Therefore, $\sup\{a_k\mid k\geq n\} = A_n\gt b-\epsilon$. This holds for all $A_n$, so $\liminf a_n = \inf\{A_n\mid n\in\mathbb{N}\} \geq b-\epsilon$. This holds for all $\epsilon\gt 0$, so $\liminf a_n \geq b$.
Now, since $b+\epsilon\gt a_n$ ultimately, then $b+\epsilon$ is an upper bound for $A_m$ all sufficiently large $m$; since the $A_m$ are decreasing, that means that $\inf A_m \lt b+\epsilon$, hence $\limsup a_n\lt b+\epsilon$; this holds for all $\epsilon\gt 0$, so $\limsup a_n \leq b$.
Now that we have established the inequalities (rather than merely asserting them), we have $b\leq \limsup a_n \leq b$, hence $b=\limsup a_n$, as claimed. QED