This is exercise $3.4.15$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $\mathcal F$ and $\mathcal G$ are nonempty families of sets and every element of $\mathcal F$ is disjoint from some element of $\mathcal G$. Prove that $\bigcup\mathcal F$ and $\bigcap\mathcal G$ are disjoint.
The author gives a proof by contradiction but I attempted to give a direct proof as follows:
Suppose $\forall A\in\mathcal F\exists B\in\mathcal G(A\cap B=\emptyset)$. Let $x$ be an arbitrary element of $\bigcup\mathcal F$. So we can choose some $A_0$ such that $A_0\in\mathcal F$ and $x\in A_0$. Since $\forall A\in\mathcal F\exists B\in\mathcal G(A\cap B=\emptyset)$ and $A_0\in \mathcal F$, we can choose some $B_0$ such that $B_0\in\mathcal G$ and $A_0\cap B_0=\emptyset$. From $A_0\cap B_0=\emptyset$ and $x\in A_0$, $x\notin B_0$. From $B_0\in\mathcal G$ and $x\notin B_0$, $x\notin\bigcap\mathcal G$. Thus if $x\in\bigcup\mathcal F$ then $x\notin\bigcap\mathcal G$. Since $x$ is arbitrary, $\forall x(x\in\bigcup\mathcal F\rightarrow x\notin\bigcap\mathcal G)$ and so $\bigcup\mathcal F\cap\bigcap\mathcal G=\emptyset$. Therefore if $\forall A\in\mathcal F\exists B\in\mathcal G(A\cap B=\emptyset)$ then $\bigcup\mathcal F\cap\bigcap\mathcal G=\emptyset$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.