Arbitrary union example

Question

In Enderton, Elements of Set Theory, an arbitrary union axiom is written in this way.

Axiom. For any set $A$, there exist a set $B$ whose elements are exactly the members of the members of $A$.
\begin{equation}
\forall x \left[x \in B \iff \left(\exists b \in A\right)x \in b\right].
\end{equation}
This set is called $\cup A$. Hence,
\begin{equation}
x \in \cup A \iff \left(\exists b \in A\right)x \in b.
\end{equation}

I have two doubts:

1. Now, suppose that $A=\{\{x\},\{x,y\}\}$. Then, $\cup A$ is the set of
   members of members of $A$. Members of $A$ are $\{x\}$ and $\{x,y\}$. Hence, $\cup A=\{x,x,y\}=\{x,y\}$. What about $\cup \cup A$? It should be the set of members of $x$ and $y$. Thereby $\cup \cup A= x \cup y$? Is it okay?

2. Suppose $A=\{x,\{x,y\}\}$. Now I'm really in trouble. What is $\cup A$?

Answer 1

If $A$ is a set with two elements, then $\bigcup A$ is the union of these two elements.

1. $\{\{x\},\{x,y\}\}$ has two elements, so $\bigcup\{\{x\},\{x,y\}\}=\{x\}\cup\{x,y\}=\{x,y\}$.
2. $\{x,y\}$ has two elements, so $\bigcup\{x,y\}$ is indeed $x\cup y$.
3. How many elements does $\{x,\{x,y\}\}$ have? Well, two: $x$ and $\{x,y\}$. Therefore $\bigcup\{x,\{x,y\}\}$ equals...