If $|I|$ < $|J|$ and $(F_i)_{i \in I}$ are finite sets, is $|\cup_{i \in I}F_i| < |J|$?

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Let $I, J$ be two infinite sets with $|I| < |J|$, and let $(F_i)_{i \in I}$ be a collection of finite sets whose union is

$$ K := \bigcup_{i \in I}F_i $$

If $s := \sup_i|F_i|$ is finite, we can give a straighforward bound for $|K|$ as follows,

$$ |K| = \left|\bigcup_{i \in I}F_i\right| \leq \left|\bigcup_{i \in I} \ \{1, \dots, s\} \times \{i\}\right| \leq |I^s| \leq |I|^s = |I| < |J| $$

My question is the following: does $|K| < |J|$ hold even when the sizes of the finite sets are unbounded?

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Yes and it follows from that fact that if $I,J$ are infinite and $F_i$ is at most countable for all $i \in I$. Then $$ \operatorname{card} \left(\bigcup_{i \in I} F_i \right) \le \operatorname{card} \left( \bigcup_{i \in I} \{i\} \times \mathbb N \right) = \operatorname{card}(I \times \mathbb N) = \operatorname{card}(I) $$ I'm using the axiom of choice here (in a couple of places, actually). It might be a worthwhile exercise to spot all the places the axiom of choice is used in this argument.