elementary set theory - Unions and Functions on Sets

Given these conditions, I seek a proof.

Let $f: A \rightarrow B$ be a function, and let $X$ and $Y$ be subsets of $A$.

Prove that $f(X \cup Y) = f(X) \cup f(Y)$.

I can't seem to figure it out. It appears obvious, but materializing a proof is troubling me. What is the best method of proof for this?

Answer

I suppose the best way to do this is to show the two inclusions (namely $f(X\cup Y)\subseteq f(X)\cup f(Y)$ and $f(X)\cup f(Y)\subseteq f(X\cup Y)$), and to use definitions. For the first inclusion, this gives :

Let $x\in f(X\cup Y)$. This means that there exists a $y\in X\cup Y$ such that $f(y)=x$. Now $y\in X\cup Y$ means that either $y\in X$ or $y\in Y$. Now if $y\in X$, then $x\in f(X)$, and if $y\in Y$, then $x\in f(Y)$. In any instance, $x\in f(X)\cup f(Y)$. We proved that any $x$ in $f(X\cup Y)$ is also in $f(X)\cup f(Y)$, that is precisely $f(X\cup Y)\subseteq f(X)\cup f(Y)$. The other inclusions proceeds similarly.