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 proof seems right, but consulting a more direct proof may help to self-test understanding, so I'll provide one below.
Since $a_n \leq b_n$ for all $n$, any upper bound on all of the $b_n$ is also an upper bound on all of the $a_n$. In particular, $\sup b_n$ is an upper bound on all of the $a_n$. By definition, $\sup a_n$ is the least upper bound on the $a_n$; setting $k = 0$, it follows that $$\sup_{ n \geq k} a_n \leq \sup_{n \geq k} b_n.$$
In other words, setting setting $A_k =\sup_{n \geq k} a_n$ and $B_k = \sup_{n \geq k} b_k$, we've shown that for $k =0$, $$A_k \leq B_k.$$
In fact, the same reasoning gives the above inequality, for all values of $k$. Taking the limit in $k$ then gives $\limsup a_n \leq \limsup b_n$, as required.
Best Answer
Your proof is fine. I only would use a different notation for the “iterated subsequence” (that is what caused my initial confusion): If $(b_{n_k})$ is a convergent subsequence of $(b_n)$ with limit $L_1$ then we can choose a convergent subsequence $(a_{n_{k_j}})$ of $(a_{n_k})$. Then $$ L_2 = \lim_{j \to \infty} a_{n_{k_j}} \ge \varliminf a_n $$ and $$ L_1 + L_2 = \lim_{j \to \infty} (a_{n_{k_j}} + b_{n_{k_j}}) \le \varlimsup (a_n + b_n) $$ so that $$ L_1 = (L_1 + L_2) - L_2 \le \varlimsup (a_n + b_n) - \varliminf a_n $$ This holds for all convergent subsequences of $(b_n)$, which implies $$ \varlimsup b_n \le \varlimsup (a_n + b_n) - \varliminf a_n \, . $$