Suppose every element of $\mathcal F$ is a subset of every element of $\mathcal G$. Prove that $\bigcup \mathcal F\subseteq \bigcap\mathcal G$. Not a duplicate of
Prove that if F and G are nonempty families of sets, then $\bigcup \mathcal F \subseteq \bigcap \mathcal G$
Validity of this proof: Prove that $\cup \mathcal{F} \subseteq \cap \mathcal{G}$
Proof that $\bigcup\mathscr F\subseteq\bigcap\mathscr G$, when every element of $\mathscr F$ a subset is of every element of $\mathscr G$
This is exercise $3.3.17$ 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 a subset of every element of $\mathcal G$. Prove that $\bigcup \mathcal F\subseteq \bigcap\mathcal G$.
Here is my proof:
Suppose $x$ is an arbitrary element of $\bigcup\mathcal F$. This means that we can choose some $A_0$ such that $A_0\in \mathcal F$ and $x\in A_0$. Let $B$ be an arbitrary element of $\mathcal G$. Since $\forall A\in\mathcal F\forall B\in\mathcal G(A\subseteq B)$, $A_0\subseteq B$. From $A_0\subseteq B$ and $x\in A_0$, $x\in B$. Thus if $B\in \mathcal G$ then $x\in B$. Since $B$ was arbitrary, $\forall B\Bigr(B\in\mathcal G\rightarrow x\in B\Bigr)$ and so $x\in\bigcap \mathcal G$. Therefore if $x\in \bigcup\mathcal F$ then $x\in\bigcap\mathcal G$. Since $x$ was arbitrary, $\forall x\Bigr(x\in \bigcup\mathcal F\rightarrow x\in\bigcap\mathcal G\Bigr)$ and so $\bigcup\mathcal F\subseteq\bigcap\mathcal G$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.

Your proof is correct! Also I like it how you justify everything. You can also check this alternative approach:

Proof. Take $A \in \mathcal F$. Then, as $A \subseteq B$ for all $B \in \mathcal G$,
$$A \subseteq \bigcap_{B \in \mathcal G} B = \bigcap \mathcal G.$$
As $A$ was arbitrary, it follows that $$\bigcup \mathcal F = \bigcup_{A \in \mathcal F} A \subseteq \bigcap \mathcal G. \quad \blacksquare$$

In the first part we used that the intersection of a non-empty family of sets is the biggest element that is contained in each element of the family, and in the second part we used that the union of a family of sets is the smallest element that contains each member of the family.
