Let $X$ be the set of Cauchy rational sequences. Let $(q_n)\in X$. Show that there exists $q\in \mathbb{Q}$ such that $(q_n)<(q)$.
Here one needs to show the following:
There exists a rational number $q$ satistfying
$(q_n)\le (q)$, which means that $0<\forall \epsilon \in \mathbb{Q}$, $\exists N\in\mathbb{N}$ such that $q_n \le q + \epsilon$ whenever $n\ge N$;
$(q_n)\not\sim (q)$, which means that $\exists 0<\epsilon_0\in\mathbb{Q}$ such that $\forall N\in\mathbb{N},\exists n\ge N_{\epsilon_0}$ satisfying $|q_n-q|\ge \epsilon_0$.
It looks like the first part is quite straightforward, but I got stuck in Part II.
Part I:
Since $(q_n)$ is Cauchy, $\forall \epsilon>0, \exists N_\epsilon \in\mathbb{N}$ such that $|q_n-q_m|\le\epsilon$ for all $m,n \ge N_\epsilon$. Fix $q_{N_\epsilon}$, then $|q_n-q_{N_\epsilon}|\le \epsilon$, which means that
$$q_n\le q_{N_\epsilon}+\epsilon$$
Part II:
Suppose there is no $\epsilon_0$ satisfying the above condition. Consider $\epsilon_0 <\epsilon$, then $N_{\epsilon_0}\ge N_{\epsilon}$. Let $n\ge N_{\epsilon_0}$, then
$$|q_n-q_{N_{\epsilon}}|\le |q_n-q_{N_{\epsilon_0}}|+|q_{N_{\epsilon_0}}-q_{N_{\epsilon}}|<\epsilon_0+\underbrace{|q_{N_{\epsilon_0}}-q_{N_{\epsilon}}|}_\text{:=k}=\epsilon_0+k$$
That's where I'm stuck. How can I complete my proof by arriving at a contradiction?