Here are what I am allowed to use: Suppose $C$ denotes the set of all rational Cauchy sequences, and ~ denotes the relation on $C$, where $\{x_n\}$~$\{y_n\}$ means their term-wise difference tends to zero. Then the quotient $C$/~ denotes all equivalent classes of $C$ under ~.
Define the $>$ to be an order on $C$/~ such that if $x,y\in C/$~ and $y>x$, then there exists $\epsilon\in \mathbb{Q}^+$ such that for all $N>0$ there is some $n>N$ and $y_n\geq x_n+\epsilon$.
Show that for all $x,y\in C/$~, and $y>x$ there is some $r\in \mathbb{Q}$ such that $y>r>x$, note $r\mapsto [r]\in C/$~, where $[r]$ is the class of constant rational sequence.
Here is my approach.
Let $x,y\in C/$~ and $y>x$, and pick $\epsilon_1>0$ such that for all $N>0$ there is some $n_1>N$ and $y_{n_1}\geq x_{n_1} +\epsilon_1$. since $\{x_n\}\in x$ and $\{y_n\}\in y$ are cauchy, the above inequality holds for all $N$, pick $N=\max\{N_x,N_y\}$($N_x,N_y$ are the $N$s where $n,m>N_x\implies |x_n-x_m|<\epsilon$,... etc), then pick $N^\prime=n_{1}+1$ then there exists some $n_2\neq n_1$ and $y_{n_2}\geq x_{n_2}+\epsilon_1$, then $$x_{n_1}<x_{n_2}+\epsilon_1 \leq y_{n_2}<y_{n_1}+\epsilon_1$$
I'm thinking we can then pick $r=x_{n_2}+\epsilon_1$, since it's fixed, and then we have $y_{n_1}+\epsilon_1>r>x_{n_1}$. This is as far as I could get, I can't seem to get rid of the $\epsilon_1$.
So here is the solution I finally came up with, the suggestion in the comment did get me to a proof, but I then figured out a way without improving that definition.
Let $x,y\in \mathbf{C}/\sim$ and $x>y$. Then for $\{x_n\}\in x$ and $\{y_n\}\in y$, there exists $\epsilon_1\in\mathbb{Q}^+$, such that for all integer $N>0$ there exists $n_1>N$ and $x_{n_1}\geq y_{n_1}+\epsilon_1$. Since this holds for all $N>0$, pick $N_1$ large enough that it satisfies the Cauchy criterion for $\{x_n\}$ and $\{y_n\}$, and let $N_2\geq n_1+1$, then there exists $n_2>n_1>N_1$ such that $x_{n_2}\geq y_{n_2}+\epsilon_1$, and we have $$|x_{n_1}-x_{n_2}|<\frac{\epsilon_1}{4}\text{ and } |y_{n_1}-y_{n_2}|<\frac{\epsilon_1}{4}$$ Define $r=\frac{x_{n_1}+y_{n_1}}{2}$ which correspond to the constant cauchy sequence $\{r\}\in\mathbf{C}/\sim$, now we have \begin{align*}r+\frac{\epsilon_1}{4}<&\frac{x_{n_2}+y_{n_2}+\frac{\epsilon_1}{2}}{2}+\frac{\epsilon_1}{4} \\\leq&\frac{2x_{n_2}-\frac{\epsilon_1}{2}}{2}+\frac{\epsilon_1}{4}=x_{n_2} \end{align*} Hence we have $r+\frac{\epsilon_1}{4}<x_{n_2}\implies \{r\}<\{x_n\}$.
With similar ways, it can be shown there exists $\epsilon\in \mathbb{Q}^+$ such that for all $N>0$ there exists $n>N$ and $y_n+\epsilon\leq r$.
Completing the proof.