Assume that $y \in f(C \cup D)$. Then $y = f(x)$ for some $x \in C \cup D$. We have $x \in C$ or $x \in D$. In the case $x \in C$ we have $f(x) \in f(C)$, so $f(x) \in f(C) \cup f(D)$. In the case $x \in D$ we also have $f(x) \in f(C) \cup f(D)$ by a similar argument. Therefore $f(C \cup D) \subseteq f(C) \cup f(D)$.

To show $f(C) \cup f(D) \subseteq f(C \cup D)$, let $y \in f(C) \cup f(D)$. We have $y \in f(C)$ or $y \in f(D)$. In the case $y \in f(C)$ we have $y = f(x)$ for some $x \in C$. In particular $x \in C \cup D$, so $y \in f(C \cup D)$. In the case $y \in f(D)$ we also have $y \in f(C \cup D)$ by a similar argument.