How do I prove this statement about the cardinalities of finite sets without using the inclusion-exclusion principle?

Question

If we were allowed to use the inclusion-exclusion principle:$|A \cup B|=|A|+|B|-|A \cap B|$ then the question statement explicitly states that the two sets are disjoint, so this proof would be trivial.

Let $A$ and $B$ be finite sets.

Assuming for this part that $A$ and $B$ are disjoint, and adopting the recursive definition of cardinality: (the empty set $\varnothing$ is finite with $|\varnothing|=0$. A set $S$ is finite with $|S|=n+1$, if there exists $s \in S$ such that $|S \backslash\{s\}|=n$ for some $n \in \mathbb{N}$. We call $|S|$ the cardinality of $S$), use induction on $|B|$ to show that $A \cup B$ is finite and that

$$ |A \cup B|=|A|+|B| \text {. } $$

Answer 1

Try an induction on $|B|$. More formally, let $P(n)$ be the statement “For all $B$ disjoint from $A$ such that $|B| = n$, $|A \cup B| = |A| + |B|$”. You should find that the inductive step and the base case follow from the two parts of the recursive definition of cardinality.

Edit: the base case $B = \emptyset$ is trivial. Now suppose $|B| = n + 1$. Then take $b \in B$ such that $|B \setminus \{b\}| = n$. Then we have $|A \cup (B \setminus \{b\})| = |A| + |B \setminus \{b\}|$ by the inductive hypothesis. And $A \cup (B \setminus \{b\}) = (A \cup B) \setminus \{b\}$. Thus, $|A \cup B| = |A| + |B \setminus \{b\}| + 1 = |A| + |B|$, as required.

Answer 2

Outline: pick any member $b$ of $B$. Then use the inductive hypothesis on $A \cup (B \setminus \{b\})$, which inductively is finite and has size $|A| + |B \setminus \{b\}|$. Now add in $b$ again.