You have the right overall idea, but you haven't actually defined a sequence or proved it converges. I think you're getting these two steps confused: the procedure you use to prove the sequence converges is a useful guide to how you want to define the sequence, but defining the sequence and proving it converges are separate tasks.
To define a sequence $(q_n)$, you need to say how to find $q_n$ if I give you $n$. You've said to pick $q_n$ such that $r-\epsilon<q_n<r+\epsilon$, but what is $\epsilon$? You need to specify how $\epsilon$ is defined in terms of $n$. Keep in mind that we don't want to talk about an arbitrary $\epsilon$ at this point--that will only come up when we are proving that $(q_n)$ converges to $r$. In order to define $q_n$ in the first place, we need to be specific about exactly what properties we are defining it to have, and so have to specify all our numbers.
Intuitively, we want the $q_n$ to be getting closer and closer to $r$ as $n$ gets larger. This suggests we want to be making your "$\epsilon$" get smaller and smaller as $n$ gets larger. There are lots of ways to do this; a simple way is to take $\epsilon$ to be $1/n$. So to be precise, we are defining $q_n$ to be some rational number such that $$r-\frac{1}{n}<q_n<r+\frac{1}{n}$$ for each $n$. We know such a $q_n$ exists by the density of $\mathbb{Q}$ in $\mathbb{R}$.
Now we need to prove that the sequence $(q_n)$ really converges to $r$. Here's where the arbitrary $\epsilon$ comes in. Let $\epsilon>0$. We want to prove that there exists $N$ such that for all $n\geq N$, $r-\epsilon<q_n<r+\epsilon$.
So, we need to somehow choose $N$ in terms of $\epsilon$. What do we need to know about $n$ in order to say that $r-\epsilon<q_n<r+\epsilon$? Well, we know that $r-1/n<q_n<r+1/n$, so as long as $1/n\leq \epsilon$, we can conclude that $r-\epsilon<q_n<r+\epsilon$. So we just need to choose $N$ to be such that $1/n\leq\epsilon$ whenever $n\geq N$. For this, we can just choose $N$ to be any integer such that $N\geq 1/\epsilon$. And that completes the argument!
I encourage you to try writing up the argument above more formally as a proof. Hidden below is how I might write it:
Let $r$ be a real number. For each $n\in\mathbb{N}$, let $q_n$ be a rational number such that $$r-\frac{1}{n}<q_n<r+\frac{1}{n}.$$ Such a rational number exists because $\mathbb{Q}$ is dense in $\mathbb{R}$. These numbers form a sequence $(q_n)$ of rational numbers which I claim converges to $r$.
To prove that $(q_n)$ converges to $r$, let $\epsilon>0$. Let $N$ be any integer such that $N\geq1/\epsilon$. Then for any $n\geq N$, $1/n\leq1/N\leq\epsilon$. We thus have $$r-\epsilon\leq r-\frac{1}{n} < q_n < r+\frac{1}{n}\leq r+\epsilon.$$ That is, for any $n\geq N$, $$|q_n-r|<\epsilon.$$ Since $\epsilon>0$ was arbitrary, this proves that $(q_n)$ converges to $r$.