This is exercise $3.4.21$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $\mathcal F$ and $\mathcal G$ are families of sets. Prove that if $\bigcup\mathcal F\nsubseteq\bigcup\mathcal G$, then there is some $A\in\mathcal F$ such that for all $B\in\mathcal G$, $A\nsubseteq B$.
The author gives a proof by contradiction but I attempted to give a direct proof as follows:
Suppose $\bigcup\mathcal F\nsubseteq\bigcup\mathcal G$. So we can choose some $x_0$ such that $x_0\in \bigcup\mathcal F$ and $x_0\notin\bigcup\mathcal G$. Since $x_0\in\bigcup\mathcal F$, we can choose some $A_0$ such that $A_0\in\mathcal F$ and $x_0\in A_0$. Let $B$ be an arbitrary element of $\mathcal G$. From $x_0\notin\bigcup\mathcal G$ and $B\in\mathcal G$, $x_0\notin B$. Since $x_0\in A$ but $x_0\notin B$, $A\nsubseteq B$. Since $B$ is arbitrary, $\forall B\in\mathcal G(A\nsubseteq B)$. From $A_0\in\mathcal F$ and $\forall B\in\mathcal G(A\nsubseteq B)$, $\exists A\in\mathcal F\forall B\in\mathcal G(A\nsubseteq B)$. Therefore if $\bigcup\mathcal F\nsubseteq\bigcup\mathcal G$ then $\exists A\in\mathcal F\forall B\in\mathcal G(A\nsubseteq B)$. $Q.E.D.$
Is my proof valid$?$
Also please help me to correct the title of my question $($I did everything I could$)$.
Thanks for your attention.