TITLE: Prove that $\mathcal{P}(A)\cup \mathcal{P}(B) \subseteq \mathcal{P}(A\cup B)$.
QUESTION [7 upvotes]: I have a presentation this Monday. I thought it was pretty straight forward but my professor wrote "You need to show why $X$ is in $\mathcal{P}(A\cup B)$, not just state that it is." I thought that I had.
Here's what I have:
Proof
Suppose $X\in\mathcal{P}(A)\cup\mathcal{P}(B)$.
By definition of union, this means $X\in\mathcal{P}(A)$ or $X\in\mathcal{P}(B)$.
By definition of power sets $X \subseteq A$ or $X \subseteq B$.
Case 1: Suppose $X \subseteq A$. Then $X \subseteq A\cup B$, and this means $X\in\mathcal{P}(A\cup B)$.
Case 2: Suppose $X \subseteq B$. Then $X \subseteq A\cup B$, and this means $X\in\mathcal{P}(A\cup B)$. 
By case 1 and 2, $X\in\mathcal{P}(A\cup B)$.
Thus $X\in\mathcal{P}(A)\cup \mathcal{P}(B)$ implies $X\in\mathcal{P}(A\cup B)$, and
therefore $\mathcal{P}(A)\cup \mathcal{P}(B)\subseteq \mathcal{P}(A\cup B). \blacksquare$
I am so bad at this. I feel so stupid.

REPLY [3 votes]: Here's an easier approach (anyway, a different one). We'll use these two general facts:
$$\text{If } X \subseteq Y \text{ then } \mathscr{P}(X) \subseteq \mathscr{P}(Y) \tag{1}.$$
$$\text{If } X \subseteq Z \text{ and } Y \subseteq Z \text{ then } X \cup Y \subseteq Z \tag{2}.$$
These are (very) easy to prove if you haven't already proved them in your course. ("Proving" them amounts to unpacking the definitions of $\subseteq, \cup$ and $\mathscr{P}(.)$, which reveals that they're trivial. In both cases, the converse is true too.) Using these truisms, the result follows simply:
Necessarily $A \subseteq A \cup B$, so by (1):
$$\mathscr{P}(A) \subseteq \mathscr{P}(A \cup B) \text{;} \tag{a}
$$
similarly, $B \subseteq A \cup B$, so again by (1):
$$\mathscr{P}(B) \subseteq \mathscr{P}(A \cup B) \text{.} \tag{b}
$$
From (a) and (b), using (2), we conclude:
$$
\mathscr{P}(A) \cup \mathscr{P}(B) \subseteq \mathscr{P}(A \cup B) \text{.}
$$