Validity of this proof: Prove that $\cup \mathcal{F} \subseteq \cap \mathcal{G}$ Here's the question as well as my thought process:
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 $\bigcup  \mathcal{F} \subseteq \bigcap  \mathcal{G}$.
$\underline{\textbf{Scratch work:}}$
Interpreting the statement "every element of $\mathcal{F}$ is a subset of every element of $\mathcal{G}$..." 
$\Rightarrow \forall A \in \mathcal{F} \forall B \in \mathcal{G} (A \subseteq B)$
While the end goal $\bigcup \mathcal{F} \subseteq \bigcap \mathcal{G}$ can be read as
$\forall x (x \in \bigcup \mathcal{F} \to x \in \bigcap \mathcal{G})$
We now have as a list of givens,


*

*$\forall A \in \mathcal{F} \forall B \in \mathcal{G} (A \subseteq B)$

*$x \in \bigcup \mathcal{F}$
and the end goal of proving 


*

*$x \in \bigcap \mathcal{G}$


Some of these expressions can be expanded


*

*$x \in \bigcap \mathcal{G} \Rightarrow \forall B (B \in \mathcal{G} \to x \in B)$

*$x \in \bigcup \mathcal{F} \Rightarrow \exists A (A \in \mathcal{F} \land x \in A)$
And we end up with a new list of givens,


*

*$\forall A \in \mathcal{F} \forall B \in \mathcal{G} (A \subseteq B)$

*$\exists A (A \in \mathcal{F} \land x \in A)$

*$B \in \mathcal{G}$
as well as a new goal to prove,


*

*$x \in B$


My final proof goes something like this:
"Suppose $B$ is an arbitrary set in $\mathcal{G}$. Suppose there is some set $A$ that is in $\mathcal{F}$ and that $x$ is an arbitrary element in that set $A$. Since every element in $\mathcal{F}$ is a subset of $\mathcal{G}$, it follows that $x$ is also an element in the arbitrary $B$ that is is $\mathcal{G}$. In other words, $x \in \bigcap \mathcal{G}$. Based on this, we can conclude that if $x \in \bigcup \mathcal{F}$ then $x \in \bigcap \mathcal{G}$. This proves $\bigcup\mathcal{F}\subseteq\bigcap\mathcal{G}$."
Does this line of thinking seem reasonable? In particular, I'm unsure about the idea that since $x$ is an element in some particular set $A$ in $\mathcal{F}$, it is then an element in every arbitrary set $B$ in $\mathcal{G}$.
I hope that this question makes sense - Working on proofs is new to me so I apologize in advance if this question seems somewhat elementary.
Thanks in advance for the help!
 A: The statement is false.
Say $\mathcal F=\{\{\{1\}\}\}$ (i.e., $\mathcal F$ has one element, and that element is $\{\{1\}\}$. 
And say $\mathcal G=\{\{1\}\}$
Then, it is true that every element of $\mathcal F$ is an element of $\mathcal G$, however
$$\bigcup \mathcal F = \{\{1\}\}\neq \{1\}=\bigcap \mathcal G$$

You say 

Interpreting the statement "every element of $\mathcal{F}$ is a subset of $\mathcal{G}$..." 
     $$\Rightarrow \forall A \in \mathcal{F} \forall B \in \mathcal{G} (A \subseteq B)$$

But that logical statement would actually translate into

Every element of $\mathcal F$ is a subset of every element of $\mathcal G$.

which is a very different statement from the original "every element of $\mathcal{F}$ is a subset of $\mathcal{G}$"


After edit
Yes, your proof is correct.

Does this line of thinking seem reasonable? In particular, I'm unsure about the idea that since $x$ is an element in some particular set $A$ in $\mathcal{F}$, it is then an element in $\textit{every}$ arbitrary set $B$ in $\mathcal{G}$.

The line of thinking is reasonable, yes.
You start with an arbitrary $B\in\mathcal G$ and an arbitrary $x\in\bigcup\mathcal F$. All you know about $B$ is that it is an element of $\mathcal G$, and all you know about $x$ is that it is an element of $\bigcup \mathcal F$.
From here on, you then take a set $A\in\mathcal F$ such that $x\in A$ because you know such a set must exist. You now have a concrete $x, A$ and $B$, and you know that $A\subseteq B$ which means that $x\in B$.
Now you take a step back and remember that $x$ and $B$ were arbitrary. So, you can conclude, that whenever you have $x\in \bigcup F$ and whenever you have $B\in \mathcal G$, you also know that $x\in B$. In other words:
$$\forall x\in\bigcup \mathcal F: (\forall B\in \mathcal G: x\in B)$$
Now you just simplify that, since $(\forall B\in \mathcal G: x\in B)$ is equivalent to $(x\in\bigcap \mathcal G)$ and you get
$$\forall x\in\bigcup \mathcal F: (x\in\bigcap\mathcal G)$$
which is what you wanted to prove.
