Use of choice function in Urysohn's lemma?

Let $$A = \mathbb Q \cap [0,1]$$. In the proof of Urysohn's lemma, we construct a family of open set such that: $$\{V_q\}_{q \in A}$$, where if $$r_1 < r_2$$, then $$V_{r_1} \subset \subset V_{r_2}$$.

The construction was made possible by some choice $$\alpha: A \rightarrow \{\text{open set}\}$$. Is there a particular reason why we chose $$A = \mathbb Q \cap [0,1]$$, i.e. a countable, dense set of $$[0,1]$$? Why can't we choose $$A = [0,1]$$? We still have a choice function $$\alpha: [0,1] \rightarrow \{\text{open sets}\}$$ such that if $$r_1 < r_2$$, then $$\alpha(r_1) \subset\subset \alpha(r_2)$$.

• How do you know that such a function $\alpha$ exists? That's not what the axiom of choice says at all (and it is inaccurate to call this $\alpha$ a "choice function"). Jun 13, 2021 at 5:05
• Ok, I misinterpreted AC, and the proof certainly does not use AC. However, the proof assumes that there exists a function $\alpha: \mathbb Q \cap [0,1] \rightarrow \text{ {open sets}}$. If there are finitely many rational numbers, $\{q_1, \cdots, q_n\}$ we may certainly construct $V_{q_1} \subset \cdots \subset V_{q_n}$. However, I am uncomfortable when I take the whole set $\mathbb Q \cap [0,1]$. (continued) Jun 13, 2021 at 5:47