Proof of indexed family set I am stuck at the following exercise in Velleman's How To Prove it:

Suppose $\{A_i\mid i \in I\}$ is a family of sets. Prove that if $\mathscr P\left(\bigcup_{i\in I}A_i\right) \subseteq \bigcup_{i\in I}\mathscr P(A_i) $, then there is some $i\in I$ such that $\forall j\in I(A_j\subseteq A_i) $. ($\mathscr P$ means powerset.)

I tried expanding this into its logical form and proceed, but its logical form has too many variables and too complicated of a structure that I couldn't get my head around it, so I turned to the solution:
i) Suppose $\mathscr P\left(\bigcup_{i\in I}A_i\right) \subseteq \bigcup_{i\in I}\mathscr P(A_i) $.
ii). Clearly $\bigcup_{i\in I}A_i \subseteq \bigcup_{i\in I}A_i$, so $\bigcup_{i\in I}A_i \in \mathscr P\left(\bigcup_{i\in I}A_i\right)$, and therefore $\bigcup_{i\in I}A_i \in \bigcup_{i\in I}\mathscr P(A_i)$.
I don't see how $\bigcup_{i\in I}A_i \in \mathscr P\left(\bigcup_{i\in I}A_i\right)$ leads to $\bigcup_{i\in I}A_i \in \bigcup_{i\in I}\mathscr P(A_i)$; in fact despite knowing its logical form $\exists i\in I\, \forall y(y\in x \to y\in A_i) $, I don't even know how to explain $\bigcup_{i\in I}\mathscr P(A_i)$ in ordinary language! 
But even if I am to ignore this, there are further trouble ahead:
iii. By the definition of union of a family, this means that there is some $i \in I$ such that $\bigcup_{i\in I}A_i \subseteq A_i$.
Does this mean $\exists i\in I \left(\bigcup_{i\in I}A_i \subseteq A_i\right)$? But where is this coming from?! It seems that the subset comes from the definition of powerset, but I don't see how $\bigcup_{i\in I}\mathscr P(A_i)$ will yield this.
iv. Now let $j\in I$ be arbitrary. Then it is not hard to see that $A_j \subseteq \bigcup_{i\in I}A_i$, so $A_j \subseteq A_i$.
For me it is vacuously true that 'it's not hard', because I can't even understand what he was doing in the prior steps.
I am a self-teaching student so I am really grateful for any help on this, thank you so much!
 A: Let’s start with $\bigcup_{i\in I}\mathscr P(A_i)$. For each $i\in I$, $\mathscr{P}(A_i)$ is the family of all subsets of $A_i$. The union of all of these families is the collection of all sets that are subsets of at least one $A_i$. That is,
$$x\in\bigcup_{i\in I}\mathscr{P}(A_i)\quad\text{iff}\quad\exists i\in I\,(x\subseteq A_i)\;.$$
Since $x\subseteq A_i$ if and only if $x\in\mathscr{P}(A_i)$, this is just an application of the fact that if $\{X_i:i\in I\}$ is any indexed family of sets, then 
$$x\in\bigcup_{i\in I}X_i\quad\text{iff}\quad\exists i\in I\,(x\in X_i)\;,$$
with $X_i=\mathscr{P}(A_i)$.
Now let’s look at $\bigcup_{i\in I}A_i \in \bigcup_{i\in I}\mathscr P(A_i)$. We know that $\bigcup_{i\in I}A_i \in \mathscr P\left(\bigcup_{i\in I}A_i\right)$, and we’re assuming that
$$\mathscr P\left(\bigcup_{i\in I}A_i\right) \subseteq \bigcup_{i\in I}\mathscr P(A_i)\;,$$
so in fact we have
$$\bigcup_{i\in I}A_i \in \mathscr P\left(\bigcup_{i\in I}A_i\right)\subseteq\bigcup_{i\in I}\mathscr P(A_i)$$
and hence
$$\bigcup_{i\in I}A_i \in\bigcup_{i\in I}\mathscr P(A_i)\;:$$
that’s just an application of the fact that if $x\in y\subseteq z$, then $x\in z$.
Now to avoid notational clutter temporarily let $a=\bigcup_{i\in I}A_i$; we’ve just seen that $a\in\bigcup_{i\in I}\mathscr{P}(A_i)$. As we saw up at the top, this means that there is some $i\in I$ such that $a\subseteq A_i$. At this point your source slips up a little, expanding $a$ to its original form and writing
$$\bigcup_{i\in I}A_i\subseteq A_i\;.$$
This is incorrect: the $i$ on the lefthand side is an index variable, a dummy variable over which we take the union, while the $i$ on the right is a specific member of $i$ such that $a\in A_i$. We can fix this in a couple of ways. One is to make it more obvious that the $i$ on the right is a specific element of $I$ by changing it to $i_0$: the fact that $a\in\bigcup_{i\in I}\mathscr{P}(A_i)$ means that there is some particular $i_0\in I$ such that $a\subseteq A_{i_0}$, i.e., such that
$$\bigcup_{i\in I}A_i\subseteq A_{i_0}\;.\tag{1}$$
The alternative would have been to change the index variable: it’s also true that $a=\bigcup_{j\in I}A_j$, so once we have a specific $i\in I$ such that $a\subseteq A_i$, we can write
$$\bigcup_{j\in I}A_j\subseteq A_i\;.$$
I’ll go with the first corrected version, with $(1)$.
Now the conclusion is pretty straightforward. For each $j\in I$ we have $A_j\subseteq\bigcup_{i\in I}A_i$ (by an easy application of the definition of the union); combining this with $(1)$, we get $A_j\subseteq A_{i_0}$. This $i_0\in I$ satisfies the conclusion of the theorem: $\forall j\in I\,(A_j\subseteq A_{i_0})$.
