Let $\{a_n\}$ be a bounded sequence. Then we define the sequences $\{a_n^+\}$ and $\{a_n^-\}$ by
$$a_n^+=\sup\{a_n,a_{n+1}\dots\}$$
$$a_n^-=\inf\{a_n,a_{n+1}\dots\}$$
We (may) then define
$$\lim a_n^+=\limsup a_n$$
$$\lim a_n^-=\liminf a_n$$
Now, you need two things to work this out:
$(1)$ Let $A$ be any bounded nonempty subset of $\Bbb R$. Then
$$\inf A\leq \sup A$$
$(2)$ Let $\{\alpha_n\}$ be a sequence such that $a_n\geq0 $ for each $n\in \Bbb N$. Then $$\lim a_n\geq 0$$
With $(1)$ you should show $$a_n^-\leq a_n^+$$ for each $n\in \Bbb N$. Monotone convergence says both $\{a_n^+\}$ and $\{a_n^-\}$ converge, since they are bounded (above/below) and are monotone (increasing/decreasing)$^{(*)}$. But
$$a_n^+- a_n^-\geq 0$$
for each $n\in \Bbb N$, so use $(2)$ to show
$$\lim a_n^+-\lim a_n^-\geq 0$$
that is:
$$\liminf a_n\leq \limsup a_n$$
$(*)$ To prove this, you need to show that if $A\subseteq B$, then $$\sup A\leq \sup B$$ $$\inf A\geq \inf B$$ Then, observe that
$$\{a_{n+1},a_{n+2},\cdots\}\subseteq \{a_n,a_{n+1},a_{n+2},\cdots\}$$
Use the definition(s) of $\limsup$ and $\liminf$. For example, the limit superior of a sequence, is defined as $\limsup a_n = \sup_{k \geq 1} \inf_{n \geq k} a_n$, and for $\liminf$ the $\sup$ and $\inf$ are switched.
Here is a better way of understanding these concepts. If $a_n$ is any sequence, then $b_n = \sup_{k \geq n} a_k$(in the extended reals) is an decreasing sequence(because the set $\{k \geq n\}$ keeps shrinking as $n$ grows larger). Therefore, every decreasing sequence either has a limit(if it is bounded) or goes to $-\infty$. The limit superior is defined as the limit of the decreasing sequence, if it exists, else it is $\pm\infty$ depending upon where the sequence goes (Note : $\limsup$ is defined over the extended reals, whence it can possibly take the values $\pm \infty$).
Similarly, the limit inferior would be defined as follows : $c_n = \inf_{k \geq n} a_k$ is an increasing sequence. So either it has a limit, which we call the limit inferior, or it goes to $\pm\infty$, whence the limit inferior would also be $\pm \infty$.
With this outlook, let us look at the questions we are given.
What is $b_n$ here? $b_n = \sup_{k \geq n} a_k$. But then $a_k$ is an increasing sequence that increases to $\infty$. So, $b_n$ is $\infty$, because the supremum is greater than each element, but the elements are themselves arbitrarily large. In particular, the limit superior is also $+\infty$. Following a similar logic gives you $\liminf a_n = -\infty$.
Of course, $a_n$ is constant, so $b_n = \sup_{k \geq n} a_k = 2$ (the supremum of the set $\{2\}$ is just $2$!) and the limit of the constant sequence $2$ is $2$. So the limit superior (analogously, limit inferior) are both equal to $2$.
- $n \sin (\frac{n \pi}{2})$.
This is a weird sequence. Look at its terms : $1,0,-3,0,5,0,-7,0,...$. In other words, this sequence can be rewritten like this : $$a_n = \begin{cases}
0 \quad n \equiv 0(2) \\
n \quad n \equiv 1(4) \\
-n \quad n \equiv 3(4)
\end{cases}$$
where $n \equiv m(k)$ means that $n-m$ is a multiple of $k$. Now, what is $b_n = \sup_{k \geq n} a_k$? If you look at the set $\{a_k,a_{k+1},a_{k+2},...\}$ then this contains a strictly increasing sequence of positive numbers, that comes from the $n$ part of the definition above. Therefore, $b_n$ is actually infinite, and so there, is the limit superior. Similarly, the $-n$ part contributes to the limit inferior becoming $-\infty$.
- $0,1,0,1,2,0,1,3,0,1,4...$
This is the trickiest of the lot. Look at $b_n = \sup_{k \geq n} a_k$. The sequence $\{a_k,a_{k+1},a_{k+2},...\}$ contains a strictly increasing sequence of positive numbers. Hence, its limit superior is $+\infty$.
On the other hand, the limit inferior? Observe the sequence $c_n = \inf_{k \geq n} a_k$. Well, observing the sequence $\{a_k,a_{k+1},...\}$, we see that $0$ must be there in this set. But then, $0$ is also the smallest element of this set! So, $c_n = 0$ for all $n$, so the limit inferior is $0$.
EDIT : I'll give you a bonus here.
The sequence $-1,-2,-3,...$ will have both $\liminf$ and $\limsup$ as $-\infty$. Try to show this from the definitions. This is a quirky point, and a nice example.
EDIT AGAIN : The new definition : the limit inferior of $x_n$ is the infimum of the set $\{v\}$ such that $v < x_n$ for at most finitely many $x_n$.
This is exactly the set $V$ of all numbers which are greater than or equal to all the numbers that appear "after some time" in the sequence i.e. there is $N$ such that for all $n > N$, $v > x_n$.
For example, lets see this for the last example. In $0,1,0,1,2,0,1,3,0,1,4,...$, what is the set $V$? Which is that element, which is greater than or equal to all elements "after some time"? I claim, that there is none of them. This is because, this sequence has a subsequence that is exactly the natural numbers, but no fixed number can be greater than all the natural numbers. So here, the limit superior is $+\infty$.
Similarly, the limit inferior is defined as the supremum over all $\{w\}$ such that $w > x_n$ for only finitely many $n$. So, what is that set here?
Well, W definitely contains all the negative integers and $0$. But anything more? No, because there are infinitely many zeros in the sequence, so anything $ a > 0$ will be greater than all those terms of the sequence which are $0$, but there are infinitely many of them so $a \notin W$! The supremum of $(-\infty,0]$ is $0$, so there you have it.
Best Answer
Here is a critique of your attempt as requested. The first part of your proof is fine given the slightly altered (but equivalent) definition $\lim\inf x_n, \lim\sup x_n$ that you are using.
However, the second part has some steps missing. You did not show that $x_n$ converges to begin with. Rather, you showed that if a subsequence is already known to be convergent, its limit must be the common value of $\lim\inf x_n = \lim\sup x_n$. This is not quite enough to conclude that $x_n$ itself the converges. There are trivial non-convergent examples like $x_n=n$ that have no convergent subsequences to begin with so they vacuously satisfy the condition "all convergent subsequences converge to the same limit". Of course, the issue with such examples can be dismissed in this case because $\lim\inf x_n = \lim\sup x_n$ exist (which crucially implies that $x_n$ is bounded) but the proof needs to be expanded to address these subtler issues.