I'm going to drop the boundedness condition (it is a legitimate difference) and use the extended reals. Also, I'm doing it for $\limsup$ ($\liminf$ is analogous).
Let $\{a_n\}_{n\geq 0}$ be a sequence in $\mathbb{R}$.
Let $E$ be the set of sub-sequential limits as in $(1)$. I will show that it is closed.
Let $x$ be a limit point of $E$. Then there are $x_n\in E$ such that $x_n\to x$ as $n\to\infty$. Each $x_n$ is the limit of a subsequence $a^n_{k_j}$.
Suppose $x\in\mathbb{R}$. For all $\epsilon>0$ there is $N$ such that $\vert x-x_n\vert<\frac{\epsilon}{2}$ if $n>N$. Furthermore, there is some index $j_n$ such that $\vert x_n-a^n_{k_{j_n}}\vert<\frac{\epsilon}{2}$. We will choose the indices $j_n$ so that $k_{j_{n+1}}>k_{j_{n}}$ (they can be chosen inductively, starting with $j_1$). Now we have $a^n_{k_{j_n}}$, a subsequence of $a_k$.
Given $\epsilon>0$, we get $N$ (since $x_n$ is convergent) and if $n>N$ we have
$\vert x-a^n_{k_{j_n}}\vert\leq\vert x-x_n\vert+\vert x_n-a^n_{k_{j_n}}\vert<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
Consequently, $x$ is a sub-sequential limit and $x\in E$.
For the cases that $x=\pm\infty$ you reword the same idea into bounding below or bounding above respectively.
So $E$ contains its limit points and is closed.
(1) is equivalent to (2)
Since $E$ is closed, $\sup E\in E$ and there is a subsequence with $\sup E$ as its limit.
Let $U=\sup E$. If there were $\epsilon>0$ such that for every $N>0$ there is $n>N$ with $a_n\geq U+\epsilon$, then there would be a subsequence with limit greater than $U$, but $U$ is the supremum of the sub-sequential limits. This proves (i).
Since there is a subsequence converging to $U$ we get (ii) by letting the $n$ be the index of an element in this subsequence.
So $\sup E$ satisfies the criteria of $U$ in (2).
(2) is equivalent to (3)
Note that $s_n$ is a weakly decreasing sequence. Since it is monotone, it has a limit (it's bounded below by $-\infty$ and we are in the extended reals).
Let $U$ satisfy the criteria in (2). By (ii), we can see that for every $\epsilon>0$, $s_m>U-\epsilon$. So $\lim s_n\geq U$.
By (i) we can see that for every $\epsilon>0$ there is $N>0$ such that $s_N\leq U+\epsilon$ for some $N$. Since $s_n$ is decreasing, we actually get $s_n\leq U+\epsilon$ for all $n>N$. So $\lim s_n\leq U$.
Consequently, $\lim s_n=U$, and we see that the definitions are equivalent.
Finally, note that since $s_n$ is decreasing, $\lim_{n\to\infty} s_n=\inf_{n\geq 0} s_n=\inf_{n\geq 0}\sup_{k\geq n}a_k$. This last expression is an alternative definition.
The only significance of the boundedness is to ensure that monotone sequences converge, working in the extended reals has the same effect with less need to separate into cases.
Best Answer
Yes, there are an infinite number of possible sub-sequences. However, thinking in terms of all possible sub-sequences is not going to help; there are just too many. If you want to find the limit points, first you should guess what they are by thinking about how the sequence behaves. Then, with guesses in hand, you should set about finding subsequences to verify your claims.
(1) $a_n=(-1)^n$: This sequence behaves quite simply. $a_1=-1$, $a_2=1$, $a_3=-1$, $a_4=1$, etc. In particular, since the sequence only ever takes the values $-1$ and $1$, these are the only possible limit points. Moreover, we can note that $a_{2k}=1$ and $a_{2k-1}=-1$ for all $k\in \mathbb N$; setting $m_k=2k$ and $n_k=2k-1$, we see that $$\lim_{k\rightarrow \infty} a_{m_k}=1,\qquad \lim_{k\rightarrow \infty} a_{n_k}=-1.$$
(2) $b_n=\sin(n \pi/2)$: like with (1), we should think about the behavior of the sequence. Plugging in the first few values of $n$, we see that $b_1=1$, $b_2=0$, $b_3=-1$, $b_4=0$, $b_5=1$; in particular, these values will just cycle. This tells us that only $1,0,-1$ can possibly be limit points; moreover, since we see those numbers crop up again and again, they must be limit points. We verify this by setting $\ell_k=2k$, $m_k=4k-3$, $n_k=4k-1$. Then $$\lim_{k\rightarrow \infty} b_{\ell_k}=0,\qquad \lim_{k\rightarrow \infty} b_{m_k}=1,\qquad \lim_{k\rightarrow \infty} b_{n_k}=-1.$$
For extra credit (and a more interesting examples):
(3) $c_n=\frac{n-1}{n}\sin(n \pi/2)$. I claim this has the same limit points as $b_n$ did. Can you prove it?