$\bigcup X$ finite implies $\mathcal P(X)$ is finite. Can anyone help with this past paper question from a Set Theory exam.

Prove that, for all sets $X$, $\bigcup X$ finite implies $\mathcal P(X)$ finite.

I am using the Kuratowski definition of finiteness, ie A is finite if every Kuratowski inductive set for A contains A.
Thanks in advance!
 A: How many subsets can $\bigcup X$ have if it is finite? Since every $A\in X$ is a subset of $\bigcup X$, this should put you on the path towards the solution.
A: $\bigcup X$ finite implies that $X$ is a finite set of finite sets (assuming a set theory without urelements). Since the powerset of any finite set is finite, the claim follows.
A: The key facts are:


*

*If $B$ is finite and $A\subseteq B$, then $A$ is finite; $\quad(1)$

*If $A$ is finite, then $\mathcal{P}A$ is finite. $\quad(2)$


Notice that $A\subseteq\mathcal{P}(\bigcup A)$. If this is not immediately obvious: fix $a\in A$, then every $\gamma\in a$ must also be in $\bigcup A$ by definition of the union; it is then clear that $$a=\{\gamma\in\bigcup A:\gamma\in a\}\in\mathcal{P}(\bigcup A).$$
Hence, by $(1)$, if $\mathcal{P}(\bigcup A)$ is finite, $A$ is finite.$\quad(3)$

Putting these facts together:
$$\bigcup X\text{ finite}\quad\stackrel{(2)}{\Rightarrow}\quad\mathcal{P}(\bigcup X)\text{ finite}\quad\stackrel{(3)}{\Rightarrow}\quad X\text{ finite}\quad\stackrel{(2)}{\Rightarrow}\quad \mathcal{P}X\text{ finite}.$$
