I came across the following problem on finite sets:
If $A_1, \dots, A_m$ is a finite list of finite subsets of a set, show that $A_1 \cup \dots \cup A_m$ is also finite.
By definition, for each positive integer $j=1, \dots, m$ there is a positive integer $r_j$ and a surjection $\sigma_j: \{1, \dots, r_j \} \to A_j$. Let $r = r_1 + \cdots + r_m$. Then we want to construct a surjection $\sigma: \{1, \dots, r \} \to A$. Now if $x \in A$, then $x \in A_1 \ \text{or} \ x \in A_2 \ \text{or} \ \dots \ \text{or} \ x \in A_m$
Note that the or is inclusive. So perhaps we can construct the surjection by saying that if $x \in A_i$ and only one $A_i$ for $i = 1, \dots, m$ then use the "individual surjections" pertaining to $A_i$. For example, if $x \in A_3$ only then there is a number $f$ in $\{1, \dots, r_3 \}$ such that $\sigma_3(f) = x$.
How do we get to the case where $x$ is in an intersection of sets? For example if $x \in A_1 \cap A_2$ then we would have to consider a surjection that involved $\{1, \dots , r_1\}$ and $\{1, \dots, r_2 \}$.