Why $P(A) \cup P(B)$ is not equivalent to $P(A \cup B)$

I know that they are equivalent if we have $\cap$ instead of $\cup$.

If we start from left, we have:

$X \in P(A) \cup P(B)$

$X \subset A \lor X \subset B$

If we start from right, we have:

$X \in P(A \cup B)$

$X \subset (A \cup B)$

And now, how do we proceed proving that they are not equivalent?

Notice that $A\cup B$ is never in $P(A)\cup P(B)$ unless $A\cup B$ is a subset of $A$ or $B$ - which never happens unless $A\subseteq B$ or $B\subseteq A$.

HINT: Let $A=\{1,2\}$, $B=\{3,4\}$. What are $P(A)$, $P(B)$ and $P(A\cup B)$?

You can see it by the cardinality of the sets. Let $A=\{a_1,a_2,\dotsc,a_n\}$ and $B=\{b_1,b_2.\dotsc,b_m\}$ be disjoint sets.

We can see that $|A|=n$ and $|B|=m$. This implies $|\mathcal P(A)|=2^n$ and $|\mathcal P(B)|=2^m$. Now $|\mathcal P(A)\cup \mathcal P(B)|=2^n+2^m-1$ (because they both share $\emptyset$). However $|A\cup B|=m+n$ and $|\mathcal P(A\cup B)|=2^{m+n}$. Therefore $\mathcal P(A\cup B) \neq \mathcal P(A)\cup \mathcal P(B)$.