Show $\mathscr{P}(A\cup B)\subseteq \mathscr{P}(A)\cup \mathscr{P}(B)$ is not true Show that it is NOT true in general that
for any sets $A$,$B$, one has $\mathscr{P}(A\cup B)\subseteq \mathscr{P}(A)\cup \mathscr{P}(B)$. 
So here is what i got:
Notice that $A\cap B \subseteq A\cup B \Rightarrow \mathscr{P}(A\cap B)\subseteq \mathscr{P}(A\cup B)$ 
Let $X \subseteq \mathscr{P}(A\cap B)$ where $X$ contains the some element(s) both A and B.
Therefore $X \nsubseteq \mathscr{P}(A\cup B) \Rightarrow \mathscr{P}(A\cup B)\nsubseteq \mathscr{P}(A)\cup \mathscr{P}(B)$
Is my proof (well disproof) correct?
Furthermore show that $\mathscr{P}(A)\cup \mathscr{P}(B) \subseteq\mathscr{P}(A\cup B)$.
Let $Y \in \mathscr{P}(A)\cup \mathscr{P}(B)$
Let $y \in Y \Rightarrow (y \in A) \lor (y \in B) \Rightarrow y\in (A\cup B) \Rightarrow Y\in \mathscr{P}(A\cup B)$
Therefore $\mathscr{P}(A)\cup \mathscr{P}(B) \subseteq\mathscr{P}(A\cup B)$
Is this ok?
 A: It is sufficient to find counterexample. For instance take $A=\{1,2\}$ and $B=\{3\}$.
A: To show it is incorrect, you have to find a set in $\mathcal{P}(A \cup B)$ that is not in $\mathcal{P}(A) \cup \mathcal{P}(B)$ for some $A, B$. It makes sense to select nonintersecting sets, and to try with small ones first. Take the sets $A = \{1\}$, $B = \{2\}$, so that:
$\begin{align}
A \cup B
  &= \{1, 2\} \\
\mathcal{P}(A \cup B)
  &= \{\varnothing, \{1\}, \{2\}, \{1, 2\}\} \\
\mathcal{P}(A)
  &= \{\varnothing, \{1\}\} \\
\mathcal{P}(B)
  &= \{\varnothing, \{2\}\} \\
\mathcal{P}(A) \cup \mathcal{P}(B)
  &= \{\varnothing, \{1\}, \{2\}\}
\end{align}$
Note that $\{1, 2\} \in \mathcal{P}(A \cup B)$ while $\{1, 2\} \not\in \mathcal{P}(A) \cup \mathcal{P}(B)$, and you are done (except for writing it up nicely).
A: Your proof is logically incorrect, because you say:

Let $X\subseteq \mathcal P(A\cap B)$ where $X$ contains some elements of $A$ and $B$

Which is wrong in two ways:


*

*If $x\subseteq\mathcal P(A\cap B)$, then $X$ contains subsets of $A$ and $B$, not elements of $A$ and $B$.

*How do you know there exists such a set that contains $A$ and $B$?



When correcting your proof, think about this hint:
If you want to prove a statement of the type "for all $A$, $B$ something is true", it is enough to find two particular sets $A$ and $B$ for which "something" is not true. 
