A trick for constructing disjoint sets

I’ve been thinking about this trick I saw this week in measure theory. Let $S_1,S_2$ be countable and suppose we are only interested in $S_1\cup S_2$. Then, without loss of generality, $S_1\cap S_2=\emptyset$.

Proof. Suppose otherwise. Clearly, $S_1\cap (S_2\setminus S_1)=\emptyset$. Now, let $\widetilde{S_2}=S_2\setminus S_1$ and hence we obtain $S_1\cup\widetilde{S_2}=S_1\cup S_2$, as required.