There is a rational number between ever 2 real numbers I have encoutered the following proof

Let $r_1<r_2$ both in $\mathbb{R}$ such that $0<r_2-r_1=\epsilon$.
  
  By Archimedean principle there is $q\in Q$ such that $0<q<r_2-r_1$.
  
  Lets look at the non-empty set of $A=\{z\in \mathbb{Z}:r_1<z\cdot q\}$ according to Archimedean it is bounded and has a minimum, $Min(A)=m$
  Therefore $r_1<m\cdot q$.
  
  All that is left to prove is that $m\cdot q<r_2 $ when $m\cdot q \in \mathbb{Q}$, let assume the contrary, $r_2\leq m\cdot q$ but $0<q<r_2-r_1$ so $q+r_1\leq r_2 \leq m\cdot q$ $\rightarrow$ $r_1<q\cdot m-q=q(m-1)$ and $m-1\in A$ which is a contradiction to $Min(A)=m$

Why does $A$ have a minimum? How does the claims add up to a contradiction to $Min(A)=m$?
 A: Formally, by the well-ordering principle, every non-empty set of positive integers contains a least element. As you correctly pointed out, this least element may not be an Integer, not even a Rational, but it will exist as a Real number, by the Well-ordering principle. But we can show this least element is an integer, using standard division $a=bq+r$.  So you assume $m$ is the minimum of $A$ and then you show that $m-1<m$ is also in $A$, so that $m-1<m$ is in $A$, contradicting that $m$ is the minimum of $A$. So all you have to show is that $A \neq \emptyset $. Consider for a minimum,  the least integer $x$ larger than $r_1/q$. Then see if $x-1$ works, to show there must be a minimum.
Consider $r_1/q$. There is an integer $N$ with $$N \leq r_1/q \leq N+1 $$
Then $N+1$ is the min of your set, since  there are no integers between $N$ and $N+1$.
Specifically, just divide $r_1$ by $q$ in the standard way, to get a quotient $b$ and a remainder $c$ with $b> c \geq 0$, so that $r_1=bq+ c$ . Then $b$ is the minimum, by definition, by properties of the Euclidean algorithm, to get $$b \leq r_1/q \leq b+1 $$. 
