True or false: for all subsets $A$ and $B$ of $X, f(A\cup B) = f(A) \cup f(B)\,$? Can somebody prove For all subsets $A$ and $B$ of $X$, $f(A \cup B) = f(A) \cup f(B)$ ?
I believe that it is true, and here is my proof.  If somebody sees something I did wrong, can you please explain?
Take any $y∈F (A∪B).$  This means $\exists x∈A∪B : f(x)=y.$  Since $x∈A∪B,$ then     $x∈A$ or $x∈B,$ so $y=f(x)∈f(A)$ or $y=f(x)∈f(B),$ so $y∈f(A)∪f(B).$ Thus   $f(A)∪f(B)⊆f(A∪B).$
Take any $y∈f(A)∪f(B),$ so $y∈f(A)$ or $y∈f(B).$  Assume $y∈f(A).$  This means $\exists x∈A : y=f(x).$  Now $A ⊆A∪B,$ so $x∈A∪B,$ so $y=f(x)∈f(A∪B).$  This shows $f(A)∪f(B)⊆f(A∪B).$
 A: Looks great. One thing to point out in the second part: you assume that $y \in f(A)$. Technically, you also need to show that the other case works as well. That is, you technically need to show that even if $y \in f(B)$, we still arrive at the same conclusion that $y \in f(A \cup B)$. The proof is exactly the same as what you did however, and so you might even get away with saying something like: "without loss of generality, we may assume that $y \in f(A)$" or "a similar argument holds for the case when $y \in f(B)$".
A: As John notes, your proof is correct, but I think it could be written more effectively. As you know, to prove $f(A\cup B)=f(A)\cup f(B)$, we must prove that $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A\cup B)\supseteq f(A)\cup f(B)$ (i.e., $f(A)\cup f(B)\subseteq f(A\cup B)$).
$(\subseteq)$: Suppose $x\in f(A\cup B)$. Thus, $x=f(y)$ for some $y\in A\cup B$. Either $y\in A$, in which case $x\in f(A)$, or $y\in B$, in which case $x\in f(B)$. Thus, in either case, $x\in f(A)\cup f(B)$. This shows that $f(A\cup B)\subseteq f(A)\cup f(B)$. 
$(\supseteq)$: Suppose $x\in f(A)\cup f(B)$. Then either $x\in f(A)$ or $x\in f(B)$. This means either that $x=f(y)$ for some $y\in A$ or that $x=f(y)$ for some $y\in B$. In either case, $x=f(y)$ for some $y\in A\cup B$, so $x\in f(A\cup B)$. This shows that $f(A)\cup f(B)\subseteq f(A\cup B)$. 
We have shown that $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A)\cup f(B)\subseteq f(A\cup B)$. By mutual subset inclusion, $f(A\cup B)=f(A)\cup f(B)$. $\blacksquare$
