Prove that $\bigcap\mathcal H\subseteq(\bigcap\mathcal F)\cup(\bigcap\mathcal G)$. Not a duplicate of
Prove that $∩\mathcal H ⊆ (∩\mathcal F) ∪ (∩\mathcal G)$.
This is exercise $3.5.17$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $\mathcal F$, $\mathcal G$, and $\mathcal H$ are nonempty families of sets and for every $A\in\mathcal F$ and every $B\in\mathcal G$, $A\cup B\in\mathcal H$. Prove that $\bigcap\mathcal H\subseteq(\bigcap\mathcal F)\cup(\bigcap\mathcal G)$.
Here is my proof:
Let $x$ be an arbitrary element of $\bigcap\mathcal H$. Now we consider two different cases.
Case $1.$ Suppose $x\in\bigcap\mathcal F$. Therefore $x\in (\bigcap\mathcal F)\cup(\bigcap\mathcal G)$.
Case $2.$ Suppose $x\notin \bigcap\mathcal F$. So we can choose some $A_0$ such that $A_0\in\mathcal F$ and $x\notin A_0$. From $\forall A\in\mathcal F\forall B\in\mathcal G(A\cup B\in\mathcal H)$  and $A_0\in\mathcal F$, it follows that $A_0\cup B\in\mathcal H$ for every $B\in\mathcal G$. Since $x\in\bigcap\mathcal H$, $x\in A_0\cup B$ for every $B\in\mathcal G$. Since $x\notin A_0$, $x\in B$ for every $B\in\mathcal G$ and so $x\in\bigcap \mathcal G$. Thus $x\in (\bigcap\mathcal F)\cup(\bigcap\mathcal G)$.
Since the above cases are exhaustive, $x\in (\bigcap\mathcal F)\cup(\bigcap\mathcal G)$. Therefore if $x\in\bigcap\mathcal H$ then $x\in (\bigcap\mathcal F)\cup(\bigcap\mathcal G)$. Since $x$ is arbitrary, $\forall x\Bigr(x\in\bigcap\mathcal H\rightarrow x\in (\bigcap\mathcal F)\cup(\bigcap\mathcal G)\Bigr)$ and so $\bigcap\mathcal H\subseteq(\bigcap\mathcal F)\cup(\bigcap\mathcal G)$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.

Your proof is okay.
It is more handsome though to prove the contrapositive statement:$$x\notin\left(\bigcap\mathcal{F}\right)\cup\left(\bigcap\mathcal{G}\right)\implies x\notin\bigcap\mathcal{H}$$

Proof:
If $x\notin\left(\bigcap\mathcal{F}\right)\cup\left(\bigcap\mathcal{G}\right)$
then some $A\in\mathcal{F}$ exists $x\notin A$ and some $B\in\mathcal{G}$
exists with $x\notin B$.
Then $x\notin A\cup B\in\mathcal{H}$ so
we conclude that $x\notin\bigcap\mathcal{H}$.
