You must first prove 2 cases:
(1) $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
(2) $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
Note that in mathematics we use the following symbols:
$\cap=$ AND = $\land$
$\cup=$ OR = $\lor$
Case 1: $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
Let $x \in A \cap (B \cup C) \implies x \in A \land x \in (B \cup C)$
$\implies x \in A \land \{ x \in B \lor x \in C \}$
$\implies \{ x \in A \land x \in B \} \lor\{ x \in A \land x \in C \} $
$\implies x \in (A \cap B) \lor x \in (A \cap C)$
$\implies x \in (A \cap B) \cup (A \cap C)$
$\therefore x \in A \cap (B \cup C) \implies x \in (A \cap B) \cup (A \cap C)$
$\therefore A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
Case 2: $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
Let $x \in (A \cap B) \cup (A \cap C) \implies x \in (A \cap B) \lor x \in (A \cap C)$
$\implies \{x \in A \land x \in B \} \lor \{ x \in A \land x \in C \}$
$\implies x \in A \land \{ x \in B \lor x \in C\}$
$\implies x \in A \land \{B \cup C \}$
$\implies x \in A \cap (B \cup C)$
$\therefore x \in (A \cap B) \cup (A \cap C) \implies x \in A \cap (B \cup C)$
$\therefore (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
$\therefore A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
The first part is good.
When you say
Therefore, $P(B) \subseteq P(A) \cup P(B).$
there is nothing "therefore" about it. $X\subseteq Y\cup X$ is basically the definition of $\cup$.
I would prefer to see some justification for why
$$
A\subseteq B\implies P(A)\subseteq P(B)
$$
It doesn't have to be long, as it's a quote obvious fact, but just a couple of words acknowledging that it isn't trivial. It is, after all, the real lynchpin of the entire argument.
Best Answer
Claim: If $A\subset C$ then $A\cup(B\cap C)=(A\cup B)\cap C$.
Proof: This is "obvious", since $A\cup(B\cap C)=(A\cap C)\cup(B\cap C)=(A\cup B)\cap C$.
Claim: If $A\cup(B\cap C)=(A\cup B)\cap C$ then $A\subset C$.
Proof: Suppose that $A\not\subset C$. Then the Left Hand Side (LHS) of our equality is also not a subset of $C$, but the Right Hand Side (RHS) is, a contradiction.