On Page 60, Set Theory, Jech(2006),
5.9 If $\{X_i : i \in I\}$ and $\{Y_i : i \in I\}$ are two disjoint families such that $|X_i| = |Y_i|$ for each $i \in I$, then $|\cup_{i \in I}X_i| = |\cup_{i \in I}Y_i|$ [Use AC]
Here's how far I goes:
$|X_i| = |Y_i|$ implies there exists a bijective function $f_i: X_i \to Y_i$ for each $i \in I$ ex ante. Since $\{X_i : i \in I\}$ is a disjoint family, for each $x \in \cup_{i \in I}X_i$, there exists exactly one $i \in I$, such that $x \in X_i$. So we could define a bijective function $f:\cup_{i \in I}X_i \to \cup_{i \in I}Y_i$, by $f(x)=f_i(x)$, if $x \in X_i$.
I don't see any usefulness of AC in problem 5.9, as opposed to problem 5.10, in which $|\cup_{i \in I}X_i| = |\cup_{i \in I}Y_i|$ is replaced by $|\prod_{i \in I}X_i| = |\prod_{i \in I}Y_i|$. The reason is that without AC, the cardinality of a cartesan product of non-empty sets is arbitary.