$\mathcal{A} \subset \mathcal{P}\left(\bigcup\limits_{X \in \mathcal{A}} X\right)$ and conditions for the opposite inclusion.

I am trying to solve the following exercise.

Let $\mathcal{A}$ be a nonempty collection of sets.

(a) Show that $$ \mathcal{A} \subset \mathcal{P}\left(\bigcup\limits_{X \in \mathcal{A}} X\right). $$ (b) Give necessary and sufficient conditions under which equality holds in part (a). Prove your assertion.

Here is my attempt. I believe I was able to solve (a), but not (b).

(a) Let $Y \in \mathcal{A}$. Then $$ Y \subset \bigcup\limits_{X \in \mathcal{A}} X $$ by definition, simply taking $X = Y$. By definition of the power set, we have $$ Y \in \mathcal{P}\left(\bigcup\limits_{X \in \mathcal{A}} X\right). $$ As $Y \in \mathcal{A}$ was arbitrary, we conclude $$ \mathcal{A} \subset \mathcal{P}\left(\bigcup\limits_{X \in \mathcal{A}} X\right), $$ as desired.

The reverse inclusion is much harder for me, and so far I have only scratchwork for it.

Let $W \in \mathcal{P}\left(\bigcup\limits_{X \in \mathcal{A}} X\right)$. Then $W \subset \bigcup\limits_{X \in \mathcal{A}} X$. If $W = X$, then the proof is complete and we have $W \in \mathcal{A}$. But there's no guarantee that $W$ must actually be one of the sets $X$. Rather, $W$ could be a union of multiple sets $X_1, X_2, X_3 \in \mathcal{A}$ (or more). I'm not sure if that itself is a sufficiently formal condition.

I don't know if "by definition, simply taking $X=Y$" is sufficient. I understand what you mean, but unless your definition of $\bigcup_{X\in \mathcal{A}} X$ is literally "the smallest set $Z$ such that $X \subset Z$ for all $X \in \mathcal{A}$", then it's unlikely to be "by definition".

In any case, it's close, and other than that step, everything looks fine.

For your second part, I think you're going about it the wrong way. You're not told to prove the equality. You're asked to find a necessary and sufficient condition for the equality to hold.

Having equality means $$\mathcal{A} = \mathcal{P}\left(\bigcup_{X\in \mathcal{A}} X\right)$$ so in particular, $\mathcal{A} = \mathcal{P}(B)$ for some set $B$. Try to show this is necessary and sufficient.