As soon as you write "Then by the definition of infimum", you are saying incorrect things.
First: if $b$ is the infimum of $A$, then for every $\epsilon\gt 0$ there exists at least one $a\in A$ such that $a\lt b+\epsilon$. However, you claim this holds for all elements of $A$, which is false. Take $A = (0,1)$. Then $\inf A = 0$; and it is indeed true that for every $\epsilon\gt0$ there exists at least one $a\in A$ with $a\lt\epsilon$; but if $\epsilon=\frac{1}{10}$, it is certainly false that every $a\in A$ is smaller than $\epsilon$.
Second: the elements of $A$ are not $a_n$s! They are suprema of infinite sequences of $a_n$s, and as such cannot be assumed to be $a_n$s. For example, if $a_n = 1-\frac{1}{n}$, then $A=\{1\}$, and no $a_n$ is equal to any element of $A$.
So that sentence is not just wrong, it's doubly wrong. The rest of course is now nonsense.
The second part does not seem to be proving anything; you are just asserting things. Why do the conditions imply the inequalities? What properties of the limit superior are you using? It's a mystery.
Rather: let $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$. Prove that $A_n$ is a decreasing sequence: $A_{n+1}\leq A_n$ for all $n\in \mathbb{N}$. Your set $A$ is precisely the set of $A_n$s.
Now let $b= \inf A = \inf\{ A_n\}$. By the definition of infimum, for every $\epsilon\gt 0$ there exists $N$ such that $b\leq A_N\lt b+\epsilon$. Since the sequence of $A_n$s is decreasing, then for all $n\geq N$ we have $b\leq A_n\leq A_N\lt b+\epsilon$, so in fact we have that $A_n\lt b+\epsilon$ ultimately. Moreover, since $a_n\leq A_m$ for all $n\geq m$, this implies that $a_n\lt b+\epsilon$ ultimately, as required.
For the second clause of the first part, let $\epsilon\gt 0$. Then $b-\epsilon\lt A_n$ for all $n$. Now remember what $A_n$ is. $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$; since $b-\epsilon\lt A_n$, there exists $k\geq n$ such that $b-\epsilon\lt a_k\leq A_n$. That is: for all $n$, there exists $k\geq n$ such that $a_k\gt b-\epsilon$, so $a_k\gt b-\epsilon$ frequently.
For part (2), let $b$ be a real number that satisfies the given properties. Since for every $\epsilon\gt 0$ we have that $b-\epsilon \lt a_n$ frequently, that means that $b-\epsilon$ is not an upper bound for $\{a_k\mid k\geq n\}$ for any $n$. Therefore, $\sup\{a_k\mid k\geq n\} = A_n\gt b-\epsilon$. This holds for all $A_n$, so $\liminf a_n = \inf\{A_n\mid n\in\mathbb{N}\} \geq b-\epsilon$. This holds for all $\epsilon\gt 0$, so $\liminf a_n \geq b$.
Now, since $b+\epsilon\gt a_n$ ultimately, then $b+\epsilon$ is an upper bound for $A_m$ all sufficiently large $m$; since the $A_m$ are decreasing, that means that $\inf A_m \lt b+\epsilon$, hence $\limsup a_n\lt b+\epsilon$; this holds for all $\epsilon\gt 0$, so $\limsup a_n \leq b$.
Now that we have established the inequalities (rather than merely asserting them), we have $b\leq \limsup a_n \leq b$, hence $b=\limsup a_n$, as claimed. QED
If we denote $A$ the set of sequential limits of a bounded sequence $a_n$ then it is proved that $\limsup a_n=\max A$.
$(1)$If $b_n \to b$ and $b_n \leq a,\forall n \in \mathbb{N}$ for some $a \in \mathbb{R}$ then $\lim_{n \to +\infty} b_n \leq a$
Now $a_n \in [-M,M]$ where $M=\sup_{n \in \mathbb{N}}$ because it is assumed bounded.
Now exists $a_{k_n} \to \limsup a_n$ because limsup is a sequential limit.
Now $a_{n_k} \in [-M,M]$ thus from $(1)$ we have that $\limsup a_n \in [-M,M]$
so $\limsup a_n=a \in [-M,M]$.
Also from this you see that limsup cannot be larger than $\sup a_n +\epsilon$ for any $\epsilon>0$
Best Answer
Rephrasing:
$(a_k){k \in \mathbb{N}}$ is is bounded.
$d_n := \sup_{k \ge n} (a_k)$ is bounded and
decreasing, hence convergent.
Let $\lim_{n \rightarrow \infty} d_n =d$.
Given:
$b> \lim \sup_{n \rightarrow \infty} (a_n)=$
$\lim_{n \rightarrow \infty} d_n=d.$
Let $0< \epsilon \le b-d.$
There exists a $n_0 \in \mathbb{N}$ s.t. for
$n \ge n_0:$
$|d_n-d| <\epsilon$, or
$d_n \lt d +\epsilon \le d + b-d=b$.
Hence
$a_n \le d_n \lt b$.