Given any two real numbers $x<y$, we can find a rational number $q$ such that $x<q<y$. Hint: use the fact that for any $x\in R, \exists N \in Z$ such that $x>1/N$ and argue by contradiction (Terence Tao-Analysis $1$).
Following the hint:
$y-x \in R => y-x > 1/N$ for some positive integer $N$. Also, since $y-x > 0 => y-x>c$ for some $c>0, c\in Z$.
Here the real number is defined by a formal symbol $LIM_{n \to \infty} a_n$, where $a_n$ is a Cauchy sequence of rationals.
We have, for every $\epsilon_0,\epsilon_1 > 0$:
$x = LIM_{n\to \infty} a_n, y = LIM_{n\to \infty}b_n$,
$i,j > N_0, |a_i-a_j| < \epsilon_0$ since $a_n$ is a Cauchy sequence
$i,j > N_1, |b_i-b_j| < \epsilon_1$ since $b_n$ is a Cauchy sequence
$n > N_2, |b_n-a_n| > c$ since $y-x>0$
Suppose there is no such $q$ that satisfies $x<q<y$. Then for every $q \in Q$ such that $q<y$ we should have that $x>q$.
But take $q = (x+y)/2$. We know that $|(x+y)/2-x|=|x/2+y/2|=|a_n+b_n|/2$. Since it is a positive real number then $|a_n+b_n|/2 > 1/N$ for some $N \in Z, N > 0$. Therefore, we get that $\exists q\in Q$ such that $q-x>0=>q>x$. Thus, we got the contradiction we wanted.