Given that $\{a_n\}$ is a sequence of $\mathbb{R}$ s.t. $a_n \leq b_n$ and $a_n \rightarrow a$ and $b_n \rightarrow b$ then $a \leq b$ 
Given that $\{a_n\}$ is a sequence of $\mathbb{R}$ s.t. $a_n \leq b_n$ and $a_n \rightarrow a$ and $b_n \rightarrow b$ then $a \leq b$

Proof Verification:
I feel I'm one step away, but can't find the right $\epsilon$
let $\epsilon > 0$, and let $b < a$. Given the convergence of $a_n$ and $b_n$ this means:
$$\forall \  \epsilon >0,\ \exists \ N_1\in \mathbb{N} \ s.t.\ \forall \ n \geq N_1 \ |a_n - a| < \epsilon$$ AND $$\forall \  \epsilon >0,\ \exists \ N_2\in \mathbb{N} \ s.t.\ \forall \ n \geq N_2 \ |b_n - b| < \epsilon$$
Consider $\epsilon = \frac{a-b}{2}$. Then:
$$ b < \frac{a}{2} + \frac{b}{2} < a_n < \frac{3a}{2} - \frac{b}{2}$$ AND 
$$\frac{3b}{2} - \frac{a}{2} < b_n < \frac{a}{2} + \frac{b}{2} < a$$
I could feel it in my soul that I have to subtract equation-2 from equation-1 and somehow manage to get a zero on one side which would show $b_n < a_n$ which would create a contradiction. But when I tried this all I got was:  $$ b-a < b_n - a_n < b-a$$  This doesn't help me because all it establishes is that $a_n \rightarrow a$ and $b_n \rightarrow b$. Which I already assumed.......What is the $\epsilon$ or the inequality I should try to get?
 A: If you want to prove an inequality in real analysis there are two main ways of doing that.
First, we could show that $a \le b + \varepsilon$ for every $\varepsilon > 0$.
The quantity $b + \varepsilon$ should remind you of the interval $(b - \varepsilon, b + \varepsilon)$. We know that $b_n \to b$ means that for every $n \ge $ some $N$, we have $b_n \in (b - \varepsilon, b + \varepsilon)$. So in particular, $b_n \le b + \varepsilon$ and therefore $a_n \le b_n \le b + \varepsilon$ (this is for $n \ge N$).
This is a step in the right direction because we have replaced $a_n \le b_n$ with $a_n \le c$ where $c = b + \varepsilon$ is constant which, at least in principle, is simpler.
Second, we could argue by contradiction, as you have done.
Suppose that $a > b$, which one should think of as $a = b + \varepsilon$ where $\varepsilon = a - b > 0$ and proceed as before. We know that for $n \ge N$ we have $b_n \in (b - \varepsilon, b + \varepsilon)$ so $b_n \le b + \varepsilon = a$ and hence $a_n \le b_n \le a$.
You will notice that this isn't particularly helpful. So we try again. What happens if we make $\varepsilon$ smaller, say to $\varepsilon = \frac{a - b}{2}$? Well, then we have
$$ a_n \le b_n \le b + \varepsilon = b + \frac{a - b}{2} \tag{$*$} $$
Again we have something of the form $a_n \le$ some constant. So if we appeal to the theorem that says that if $a_n \le c$ (a constant) and $a_n \to a$ then $a \le c$ we have from $(*)$:
$$ a \le b + \frac{a - b}{2} < b + (a - b) = a, $$
and this is a contradiction.
As an exercise, you might like to finish the first method as well.
A: First of all, the statement as it was written at the beginning is not true because you can take $a_n$ and $b_n$ be the same sequence as a counterexample. Perhaps you mean $a\leq b$, then I can give an easy proof. Let $\epsilon>0$. There exists a large enough $n$ which satisfies both $a-\frac{\epsilon}{2}<a_n$ and $b_n<b+\frac{\epsilon}{2}$. For this specific $n$ we have:
$a<a_n+\frac{\epsilon}{2}\leq b_n+\frac{\epsilon}{2}<b+\frac{\epsilon}{2}+\frac{\epsilon}{2}=b+\epsilon$
So $a<b+\epsilon$. This is true for any $\epsilon>0$. But if we assume $a>b$ then we get it isn't true for $\epsilon=\frac{a-b}{2}$ which is a contradiction. Hence $a\leq b$. 
A: New answer:
Suppose not! If $a\not\le b$ then $b<a$ so $0<a-b$.  Let $c=a-b$, then $c>0$.  Then $c=\lim_{n\to \infty} a_n - \lim_{n\to \infty} b_n = \lim_{n\to \infty} (a_n-b_n) = \lim_{n\to \infty} c_n$
Now by the definition of "limit", it follows that for 

for all $\varepsilon>0$, there exists $N$ such that for all $n\ge N$, we have $|c_n-c|<\varepsilon$

Why does this fail?
Well... $c>0$ is always positive and $c_n$ is always negative.  Can you finish it from there?

Original answer (for $a<b$ case):
Let $a_n=b_n=1$ (for every $n$) then $a_n\le b_n$ and $\lim_{n\to \infty} a_n=a=1$ and $\lim_{n\to \infty} b_n=b=1$ but $a\not<b$
A: From what you have done follows than for $n \geqslant N_1, N_2$:
$$b_n < \frac{a}{2} + \frac{b}{2} < a_n$$
so $b_n < a_n$, which is a contradiction.
