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.
You proof is not correct. Since $\varepsilon>0$, you clearly cannot have $M-\varepsilon>M$.
Take $\varepsilon>0$. Then $M+\varepsilon>M$ and therefore $M+\varepsilon$ is not a lower bound of $\{x_n\mid n\in\mathbb N\}$. Therefore, $x_N<M+\varepsilon$, for some $N\in\mathbb N$. Since the sequence is monotonic and decreasing, $n\geqslant N\implies x_n<N+\varepsilon$ too. So$$n\geqslant N\implies x_n\in[M,M+\varepsilon)\implies\lvert x_n-M\rvert<\varepsilon.$$
Best Answer
Perhaps a detail that can be added :
$t_n=\sup \{x_k| k\ge n\}$.
$t_{n+1}=\sup \{x_k|k\ge n+1\}$.
1) Assume $x_n \ge x_k$ for $k \ge n+1$, i.e.
$x_n$ is an upper bound for $\{x_k|k \ge n+1\}$.
Then $t_{n+1} =\sup \{x_k|k\ge n+1\} \le x_n$,
since $t_{n+1}$ is the least upper bound for $\{x_k| k \ge n+1\}$.
Obviously:
$x_n \ge x_k$, $k \ge n$, and we have
$x_n= t_n$;
Altogether
$t_{n+1} \le t_n$.
2) Assume $x_n \lt x_k$, for $k \ge n+1$.
Then
$t_n=t_{n+1}$, and we are done