How to show if $a$$\leq$$b_1$, for every $b_1>b$, then $a$$\leq$$b$ where a,b$\epsilon$R? Not positive on the proper approach to this problem. 
My first thought:
$a $ $\leq$ $b_1$ means either $a=b_1$ or $a<b_1$. 
Should it broken up into cases? 

Second attempt: 
Assume, to the contrary, $a > b$, then let $b_1$= $(a+b)/2$. It follows that, $a > b_1 > b$, which is a contradiction. 
Thus, $a\leq b$. $ \Box$$ $
 A: What would happen if, to the contrary, $a>b$?
A: I think there's a mistake in the question.  Play with each case 
a = c and a 
Note- my suggestion is not proof by contradiction which assumes the conclusion false, then derives a contradiction => the assumption is invalid.
A: If $a>b$, then let $c:=\frac{a+b}{2}$, and you have $a> c >b$, which is a contradiction.
Therefore $a \leq b$.
A: Let $a$ be a real number. If there is a real $b$ such that $a \leq c$ for every real $c > b$ and $a > b$, then there is a real $c$ such that $b < c < a;$ a contradiction. 
A: Ok, you changed the letters on us, which creates confusion with existing answers.  The question I'm answering is

How to show if a≤b1, for every b1>b, then a≤b where a,bϵR?

Let a=b1,  then by the hypothesis    a > b.
but this contradicts the desired conclusion a≤b
Notice I assumed nothing except for a specific case allowed in the original statement.  So as given the theorem is false.  (A single counterexample can prove an entire theorem false)
Is there something I missed?
