A number $a \in \mathbb R$ is called singular if for all $\epsilon > 0$, there exists $Q_\epsilon$ such that for all $Q \ge Q_{\epsilon}$, there exist integers $p \in \mathbb Z$ and $q \in Z$ such that
\begin{equation} \|aq+p\|\le \epsilon Q^{-1} ~\text{and}~ 0<\| q \| \le Q \end{equation}
I want to prove that a singular number is necessarily rational.
My proof: Suppose $a$ is irrational. Recall Hurwitz's theorem, in which $\frac{1}{\sqrt{5}}$ is optimal. We have for any $\epsilon < \sqrt{5}$, if we let $a=(1+\sqrt{5})/2$, there exists only finitely many relatively prime integers $p,q$ such that $|aq+p|\le \epsilon/q$. Namely there are infinitely many relatively prime integers $p,q$ such that $$|aq+p|>\epsilon/q.$$ Namely for all $Q_{\epsilon}>0$, we can always find a large $q\ge Q_\epsilon$ such that $|aq+p|>\epsilon/q$. This means that $a$ is not a singular number.
Is my proof correct? I feel like I have used a very strong result to prove a basic thing. If there are any other proofs that doesn't need to optimality of the Hurwitz's theorem, please let me know.