Proving $(\bigcup F)\setminus (\bigcup G)\subseteq \bigcup (F\setminus G)$ 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.
 A: Just for fun, here is your own correct proof in a calculational format.  This makes the $\bigcup$- quantifications explicit, which prevents mistakes.
For every $\;x\;$,
\begin{align}
& x \in (\bigcup F) \setminus (\bigcup G) \\
\equiv & \qquad \text{"definition of $\;\setminus\;$"} \\
& x \in \bigcup F \;\land\; x \not\in \bigcup G \\
\equiv & \qquad \text{"definition of $\;\bigcup\;$, twice"} \\
& \langle \exists A : x \in A : A \in F \rangle \;\land\; \lnot \langle \exists A : x \in A : A \in G \rangle \\
\equiv & \qquad \text{"logic: DeMorgan on right hand side"} \\
& \langle \exists A : x \in A : A \in F \rangle \;\land\; \langle \forall A : x \in A : A \not\in G \rangle \\
\equiv & \qquad \text{"logic: use right hand side in left hand side quantification"} \\
& \langle \exists A : x \in A : A \in F \land A \not\in G \rangle \;\land\; \langle \forall A : x \in A : A \not\in G \rangle \\
\Rightarrow & \qquad \text{"logic: weaken -- we don't need the right hand side anymore"} \\
& \langle \exists A : x \in A : A \in F \land A \not\in G \rangle \\
\equiv & \qquad \text{"definition of $\;\setminus\;$"} \\
& \langle \exists A : x \in A : A \in F \setminus G \rangle \\
\equiv & \qquad \text{"definition of $\;\bigcup\;$"} \\
& x \in \bigcup (F \setminus G) \\
\end{align}
From this the theorem follows by the definition of $\;\subseteq\;$.
