Suppose $F$ and $G$ are families of sets. Suppose $x$ is an arbitrary element of $(\bigcup F)\setminus (\bigcup G)$.
It follows that $x\in \bigcup F$ and $x\not \in \bigcup G$.
Since x∈∪F, therefore there exists A∈F such that x∈A. (1)
Since x∉∪G, x can not be an element of any set in G.
Since x∈A, so A∉G.
Since A∈F and A∉G, so A∈F\G.
Since x∈A and A∈F\G, therefore x∈∪(F\G).
Since x is arbitrary, we can conclude that (∪F)\(∪G)⊆∪(F\G).
Above is my initial attempt on proving it. But starting from (1), I think I have made a mistake. Since the $x$ mentioned above is not an arbitrary element of the chosen $A\in F$, we cannot straightforward conclude that $A\not \in G$ since $x$ is only an arbitrary element of $\left(\bigcup F\right)\setminus \left(\bigcup G\right)$. Perhaps there exists some element in $A$ which is an element of some set in $G$. I am stuck here; I have absolutely no idea about solving it. Please help. Thanks in advance.