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.
Your reasoning seems right to me, but conceptually speaking I believe the proof is much more easy.
Proof from Rudin's definitions:
Put $t^*=\limsup_{n\to\infty} t_n$ and $s^*=\limsup_{n\to\infty}s_n$. For a contradiction, suppose that $s^*>t^*$ and put $\epsilon=\frac{s^*-t^*}{2}, x=s^*-\epsilon$. Notice that $x>t^*$.
With the definitions as in Rudin's book, by Theorem 3.17(2), there is $N_0\in\mathbb{N}$ such that for all $n\geq N_0$ we have $t_n<x$. On the other hand, by Theorem 3.17(1), $s^*\in E$ and so there is a subsequence $s_{n_k}$ that converges to $s^*$. The last implies in particular that there is $n_k\geq \max\{N_0,N\}$ such that $s_{n_k}\in (s^*-\epsilon,s^*+\epsilon)$, and we would have $$s_{n_k}>s^*-\epsilon=x>t_{n_k},$$ contradicting that $s_n\leq t_n$ for $n\geq N$.
To prove that $\liminf_{n\to\infty} s_n\leq \liminf_{n\to\infty}t_n$ you have two options: either you proof the analog of Theorem 3.17 for $s_*$, or you show that for a sequence $\{x_n\}$, $$\liminf_{n\to\infty} x_n=\limsup_{n\to\infty} (-x_n).$$
Proof with other definition:
First, note that the following:
Claim: If $\{s_n\},\{t_n\}$ are sequences such that $s_n\leq t_n$ then $$ \inf_n s_n\leq \inf_n t_n,\text{ and, }\sup_n s_n\leq \sup_n t_n.$$
- Proof of the Claim: (If you don't need it, you can scroll down). Let us show first that $\inf_n s_n\leq \inf_n t_n$. Suppose for a contradiction that $\alpha=\inf_n s_n > \inf_n t_n=\beta$, and consider $\epsilon=\alpha-\beta>0.$ By definition of inf, there is an element $t_n$ such that $$t_n\in (\beta,\beta+\epsilon)\Rightarrow\beta\leq t_n<\beta+\epsilon=\alpha\leq s_n\Rightarrow t_n<s_n,$$ contradicting the hypothesis that $s_n\leq t_n$. A similar proof can be done to show that $\sup_n s_n\leq \sup_n t_n$.
From this claim, we can prove that $\liminf_{n\to\infty}s_n\leq \liminf_{n\to\infty}t_n$ as follows:
Remember that $\liminf_{n\to\infty}s_n=\sup_{N\in\mathbb{N}}\inf_{n\geq N} s_n$, and similarly for $\liminf_{n\to\infty} t_n$. For $k\geq N$, consider the sequences given by $\sigma_k=\inf_{n\geq k}s_k$ and $\tau_k=\inf_{n\geq k}t_k$.
Since $s_n\leq t_n$ for every $n\geq N$, we have by the Claim that $\sigma_k\leq \tau_k$ for every $k\geq N$, and again by the claim,
$$\liminf_{n\to\infty} s_n=\sup_{k}\sigma_k\leq \sup_k \tau_k=\liminf_{n\to\infty}t_n.$$
A similar proof can be done from the definition of $\limsup$ changing a little bit the definition of the sequences $\sigma_k,\tau_k$, to finally show that $$\limsup_{n\to\infty}s_n=\inf_{k}\sigma_k\leq \inf_k\tau_k=\limsup_{n\to\infty} t_n.$$
Best Answer
We first construct a sub-sequence in $S$ that has limit $\lim_{n\to\infty} \sup_{k>n} s_k$. We do this by observing that there's a $l(n)>n$ such that $\sup_{k>n}s_k \ge s_{l(n)} \ge \sup_{k>n}s_k-2^{-n}$, now $\lim_{n\to\infty}s_{l(n)} = \lim_{n \to \infty} \sup_{k>n} s_k$.
Now as $l(n)>n$ we have $l(n)\to\infty$ so $l(n)$ has an increasing subsequence $t_n$ and by that we can construct the desired subsequence of $s_n$.
That is we have proved that
$$\sup S \ge \lim t_n = \lim_{n\to\infty}\sup_{k>n}s_k$$
To prove the opposite lets consider "synchronized" subsequence $u_n$ (that $u_n=s_n$ or $u_n=u_{n-1}$ - just repeating the elements in the subsequence to keep them in sync) - this kind of construct doesn't alter the definition if $S$. Now $\{u_k: k>n\}\subseteq \{s_k: k>n\}$ and therefore $\sup_{k>n}u_k \le sup_{k_n}s_k$.
This means that:
$$\lim_{n\to\infty} u_k \le \lim_{n\to\infty}\sup_{k>n}u_k \le \lim_{n\to\infty}\sup_{k>n}s_k$$
That is for all subsequences in $S$ the limit is less than or equal the $\liminf$ of definition (2) that is:
$$\sup S\le\lim_{n\to\infty}\sup_{k>n}s_k$$