Is this a correct proof that there are infinitely many irrationals between two rationals?
Take $n\in \mathbb N$, $q=\cfrac 1n \in \mathbb Q$.
I chose this because $1$ is the smallest positive integer and we want to take $q \in \mathbb Q$ as small as possible. As $n \to \infty, q \to 0$ so for any $\epsilon \gt 0$, we can find $q$ with $\epsilon \gt q \gt 0$.
Now take an irrational number $1 \gt \alpha\gt \epsilon$.
Consider $\frac 1 n \alpha$. This is obviously an irrational number always less than $q$. Since there are infinitely many irrational numbers $\alpha \in (\epsilon, 1)$ there are infinitely many irrationals between $0$ and $\epsilon$ in $\mathbb Q$.