I used the axiom of choice to solve the following Problem 3-13 in "Calculus on Manifolds" by Michael Spivak.
Can I avoid using the axiom of choice to solve 3-13?
Problem 3-13.
(a) Show that the collection of all rectangles $[a_1,b_1]\times\cdots\times [a_n,b_n]$ with all $a_i$ and $b_i$ rational can be arranged in a sequence.
(b) If $A\subset\mathbb{R}^n$ is any set and $\mathcal{O}$ is an open cover of $A$, show that there is a sequence $U_1,U_2,U_3,\dots$ of members of $\mathcal{O}$ which also cover $A$. Hint: For each $x\in A$ there is a rectangle $B=[a_1,b_1]\times\cdots\times [a_n,b_n]$ with all $a_i$ and $b_i$ rational such that $x\in B\subset U$ for some $U\in\mathcal{O}$.
(a) Since $\mathbb{Q}^{2n}$ is countably infinite, the collection of all rectangles $[a_1,b_1]\times\cdots\times [a_n,b_n]$ with all $a_i$ and $b_i$ rational can be arranged in a sequence.
(b) Let $B_1,B_2,B_3,\dots$ be a sequence of all rectangles $[a_1,b_1]\times\cdots\times [a_n,b_n]$ with all $a_i$ and $b_i$ rational.
If $x\in A$, there is an element $U\in\mathcal{O}$ such that $x\in U$ because $\mathcal{O}$ is a cover of $A$. Since $U$ is open and $\mathbb{Q}$ is dense in $\mathbb{R}$, there is $i\in\{1,2,3,\dots\}$ such that $x\in B_i\subset U$. Let $I=\{i\in\{1,2,3,\dots\}:\text{ there is }U\in\mathcal{O}\text{ such that }B_i\subset U\}=\{i_1,i_2,i_3,\dots\}$. Let $i\in I$ and $\mathcal{O}_i=\{U\in\mathcal{O}:B_i\subset U\}$. $\mathcal{O}_i\neq\varnothing$. By the axiom of choice, there is a mapping $V:I\to\mathcal{O}$ such that $V_i\in\mathcal{O}_i$ for $i\in I$. Let $U_j=V_{i_j}$. $\{U_1,U_2,U_3,\dots\}\subset\mathcal{O}$ is a cover of $A$.