Question on Proof Validity: Arbitrary Union of Power Set In Exercise 2.24 of Enderton's Elements of Set Theory, we are asked to show that $$\bigcup{\{\mathcal{P}}X \ \mid X \in A\} \subseteq \mathcal{P} \bigcup A.$$ 
Fascinated by this exhilerating problem, I started off my proof by taking an arbitrary element $x$ of $\bigcup{\{\mathcal{P}}X \ \mid X \in A\}$ and showing that by definition of the arbitrary union, there exists a set $b \in \mathcal{P} X$ such that $x \in b$. I then proceeded to  show that because $b \in \mathcal{P} X$ implied $b \subseteq X$, that $x \in X$, and that because $X \in A$, that $x \in \bigcup A$. 
But as one can probably guess, this is not enough. I must show that $x \subseteq \bigcup A$. Yet, usually, it is not the case that if $a \in B$ and $B \subseteq C$ that $a \subseteq C$. And so I am left sad, confused, and a bit upset, because I can't seem to figure out how to come to that conclusion.
To add to my confusion, looking up a solutions manual online, I find a proof that assumes just that, claiming that "it follows that $x \in X$, but $X \subseteq \bigcup A$, so $x \subseteq \bigcup A$". This solution is probably right, and there's probably some trick involved, but I have no idea what the heck that trick is and I would greatly appreciate any help I can get.
As a poor little senior in high school trying to self-study random maths in quarantine, please help a fellow out!
 A: You are applying the definition of union incorrectly. $$ x\in \bigcup\{\mathcal PX:X\in A\} $$ means that $x\in \mathcal P X$ for some $X\in A,$ not that $x\in b$ for some $b\in \mathcal P X$ for some $X\in A.$
A: Let’s see if I can explain the source of your confusion. 
By definition, if $Z$ is a set, then
$$\bigcup Z = \{b\mid \exists W\in Z(b\in W)\}.$$
You are correct on that.
Here what we have is
$$\bigcup \{\mathcal{P}X\mid X\in A\}.$$
The set $Z$ from the definition is
$$Z=\{\mathcal{P}X\mid X\in A\}.$$
That means that the sets $W$ in the description of $\bigcup Z$ will be the sets $\mathcal{P}X$ with $X\in A$.
So by definition, the union you want consists of all elements $b$ for which there exists an element $W$ of $\{\mathcal{P}X\mid X\in A\}$ with $b\in W$. But that means that $W$ must be of the form $\mathcal{P}X$ with $X\in A$. In other words, you are looking for all elements $b$ such that there exists $X\in A$ with $b\in\mathcal{P}X$. Or in yet other words, there exists $X\in A$ with $b\subseteq X$.  
Instead, you seem to be looking at elements of $$\bigcup\left(\bigcup\{\mathcal{P}X\mid X\in A\}\right).$$
That is, you went one level too deep.
The confusion seems to arise because here they are describing the set in the unary union by its elements rather than by its name. 
A: Thanks to the wonderful comments of @Arturo Magidin and @spaceisdarkgreen, I have figured out both the problem with my original proof, along with the solution to this exercise.
My original answer to the problem was based on a misunderstanding of the notation $\bigcup \{ \mathcal{P} X \mid x \in A\}$. I had assumed that a member of this union would be a member of $\mathcal{P} X$ for some $X \in A$. This is incorrect. Instead, I seemed to be looking at the elements of the union of that union. 
To show my thanks, I would like to upload the completed proof. I'll also answer the second part of the original question which asks under which conditions equality holds.
Proof: Let us take an arbitrary element $x \in \bigcup \{ \mathcal{P} X \mid x \in A\}$. Then $\left( \exists b \in \{\mathcal{P} X \mid X \in A\}\right) x \in b$. In other words, $\exists b = \mathcal{P} X$ for some $X \in A$ with $x \in \mathcal{P} X$. This implies that $x \subseteq X \subseteq \bigcup A$, which by definition of the power set, implies that $x \in \mathcal{P} \bigcup A$. 
Note: $X \subseteq \bigcup A$ because for every element of $X$, there is an element in $A$, mainly $X$ that holds it.
To examine under which conditions equality holds, let us take another arbitrary element $x \in \mathcal{P} \bigcup A$. By definition of the power set, $x \subseteq \bigcup A$. As such, for $x \in \bigcup \{ \mathcal{P} X \mid x \in A\}$ to hold, $x$ must be a subset of $X$ for some $X \in A$, meaning that $\bigcup A$ must be a subset of $X$ for some $X \in A$ for equality to hold.
