Functions-Set Theory Proof that $f(C cup D) = f(C) cup f(D)$

I'm revisiting set theory and am troubled by this question.

Let $f:A \rightarrow B$, and $C \subset A$, $D \subset A$.

Prove that $f(C \cup D) = f(C) \cup f(D)$.

Any thoughts?

Answer

I'll show $\subseteq$. Let $y\in f(C\cup D)$. Then there exists an $x\in C\cup D$ such that $f(x)=y$. This means $x\in C$ or $x\in D$, hence $f(x)\in f(C)$ or $f(x)\in f(D)$. This implies $f(x)\in f(C)\cup f(D)$ and we've established $$f(C\cup D)\subseteq f(C)\cup f(D).$$ Approach the other containment in a similar manner.