Why is AC needed for $|bigcup X_i|=|bigcup Y_i|$, $forall i$ $|X_i|=|Y_i|$, ${X_i}_{ iin I}$, ${Y_i}_{iin I}$ pairwise disjoint?

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.