While looking at this problem, I was thinking about the more general statement:
For all sets $I,X$ and $U:I\to \mathcal P(X)$, the power set of $X$, there exists a $V:I\to \mathcal P(X)$ with the property that for all $i\in I$, $V_i\subseteq U_i$, for all $i,j\in I, i\neq j: V_i \cap V_j=\emptyset$ and $$\bigcup_{i\in I} U_i = \bigcup_{i\in I} V_i$$.
Obviously, if you have the axiom of choice, you can well-order $I$, and you get this result by the reasoning in the answer to that other question. But does this result imply choice, or is it weaker?
If it does imply choice, it seems like perhaps the most direct approach would be to show it implies Zorn's lemma. But I'm struggling to prove one way or the other.
They are equivalent. Suppose that $\{ X_i \}_{i \in I}$ is a family of nonempty pairwise disjoint sets and let $X = \bigcup_{i \in I} X_i$. For each $x \in X_i$ define $A_x = \{ i \}$. Now take a pairwise disjoint family $\{ B_x \}_{x \in X}$ with $B_x \subseteq A_x$ and $\bigcup_{x \in X} B_x = \bigcup_{x \in X} A_x = I$. For each $i \in I$ pick the unique $x \in X_i$ such that $B_x \neq \emptyset$. This is your choice function for $\{ X_i \}_{i \in I}$.