I was reading Wilcox and Myers' "Introduction to Lebesgue Integration and Fourier Series" (Dover). On p.18, the authors try to prove that every non-empty open subset $G$ of the reals can be expressed uniquely as a countable union of pairwise disjoint open intervals. Their proof is quite similar to the one in this question or the one in Steven Krantz's "A Guide to Topology" (MAA). So, I guess this is a rather usual proof.
At some point, the authors have shown that when $G$ is bounded, $G$ can be written as a disjoint union of non-empty open intervals $\cup_{x\in G}I_x$, where either $I_x\cap I_y=\phi$ or $I_x=I_y$ for any $x,y\in G$. Then they write:
That this collection is countable follows from the fact that one can choose a rational number in each $I_x$ (using the axiom of choice), and the disjointness of the intervals guarantees that no duplication will occur in the choice of rationals.
Why do we need the axiom of choice? If we denote $I_x=(a_x,b_x)$ and define \begin{align} n_x&=\left\lceil\frac1{b_x-a_x}\right\rceil+1,\\ q_x&=\min\left\{\frac{j}{n_x}\in I_x:\ j\in\mathbb Z\right\}, \end{align} doesn't it suffice to pick $q_x$ as a representative of $I_x$? What am I missing?
