Define an equivalence relation on Cauchy sequences in $\mathbb{Q}$ by $(x_n)\equiv (y_n)$ if $\lim (x_n-y_n)=0$.
The question I am trying to prove is that
If $(x_n)$ and $(y_n)$ are Cauchy sequences in $\mathbb{Q}$ such that $(x_n)(y_n)\equiv (0)$ then $(x_n)\equiv (0)$ or $(y_n)\equiv (0)$.
Here, I am considering the problem only over $\mathbb{Q}$ and not considering anthing about $\mathbb{R}$.
Some part of proof: Suppose $(x_n)\not\equiv (0)$. We show that $(y_n)\equiv (0)$.
Since $(x_n)\not\equiv (0)$, so $(x_n)$ is a Cauchy sequence but its limit is not $0$.
Hence there is an $\epsilon>0$, such that for every $i\in\mathbb{N}$, there is $n_i$ such that $|x_{n_i}|\ge \epsilon$.
What this says is that infinitely many terms of $(x_n)$ are bounded away from $0$.
How to proceed now, to show that $(y_n)\equiv (0)$?
Correct. You can use that and the fact that $(x_n)$ is a Cauchy sequence to get the even stronger statement that there exists $N\in\mathbb{N}$ such that $|x_n|\geq\frac{\epsilon}{2}$ for all $n\geq N$. That makes it straightforward to show $y_n\rightarrow 0$.