Proof verification that if $a_n\leq b_n$ then $\limsup a_{n} \leq \limsup b_{n} $ 
Suppose $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are sequences such that for every $n, a_{n} \leqslant b_{n} .$ Prove that
  If $a_n\leq b_n$ for all $n$ then $\limsup a_{n} \leq \limsup b_{n} $

(proof)
Let $ A = \limsup a_{n} $ and let $ B=\limsup b_n $. 
First of all, I'm aware that there are many questions like this on the site, but they all seem to be related to either $\limsup$ or $\liminf$ and I couldn't find anything that would help me with my problem. I've done some Googling and found some great resources, but I'm still not quite sure how to get to some steps and would like your assistance.
The problem is as follows:
Assume to the contrary that $ B<A $ and that for all $ n\in\mathbb{N}, a_n\leq b_n. $
We know there is a subsequence $ \{a_{n_k}\} $ that converges to $ A $.   Let $ \epsilon = \frac{B-A}{2} $. 
Then we know that there is a $ K $ such that for $ k>K, |a_{n_k}-A|<\epsilon$. 
Given $a_n < b_n$ prove that $\lim_{n\to \infty}(a_n) \le \lim_{n\to\infty}(b_n)$. The proof is then done by contradiction, assuming that $a = \lim_{n\to \infty}(a_n) > b =\lim_{n\to\infty}(b_n)$.
We take an $\epsilon = \frac{a-b} 2$, so that the $\epsilon$-neighborhoods of $a$ and $b$ are disjoint. From the definition of limits, we now know that there is such a $N$, so that $\forall n > N : |a_n-a|<\frac\epsilon2$ and $|b_n-b|<\frac\epsilon2$.
The next step is absolutely always confusing. Two variants I've found are either:
We know that there are infinity many terms of $ a_n $ in $ (A-\epsilon , \epsilon +A) $. 
$a_n>a-\epsilon=a-\left(\frac{a-b} 2\right)=b+\left(\frac{a-b} 2\right)=b+\epsilon>b_n$
However for this same $ \epsilon $ we know that there are only finitely many terms of $ b_n $ greater than $ B+\epsilon $. 
Finding the maximum subscript $ n $ of these finitely many $ b_n $ greater than $ B+\epsilon $ gives us a corresponding value of $ N $ such that for $ n>N, b_n < B+\epsilon$. 
Let $ M=\max\{K,N\} $. 
Then it follows that for $ n>M, b_n<B+\epsilon$ and $ a_n > A-\epsilon
= B+\epsilon $. 
So we have found an $ a_n > b_n.  $ Contradiction.
I am wondering if there is anything wrong with my proof
Edit:
Definition: Let $\left\{a_{n}\right\}$ be a sequence of real numbers. Then $\lim$
$\sup a_{n}$ is the least upper bound of the set of subsequential limit points of
$\left\{a_{n}\right\},$ and $\lim \inf a_{n}$ is the greatest lower bound of the set of subsequential
limit points of $\left\{a_{n}\right\} .$
 A: 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.
A: Option:
$\limsup_{n \rightarrow \infty}x_n:=\lim_{n\rightarrow \infty}(\sup{x_k| k\ge n});$
$c_n:=\sup ${$a_k| k\ge n$};  
$d_n:=\sup ${$b_k| k \ge n$};
$(\star)$ $c_n \le d_n$, since $a_k \le b_k$, $k \in \mathbb{N}.$
Then 
$\limsup_{n\rightarrow \infty} a_n=\lim_{n \rightarrow \infty}c_n \le \lim_{n \rightarrow \infty} d_n =\limsup_{n \rightarrow \infty}b_n.$
P.S.  As a not so difficult exercise prove $(\star)$.
