My try:
Suppose $x \in (A \cup B) \cap (A \cup C)$. We know $x \in A \cup B$ and $x \in A \cup C$. We know $x \in A$ or $x \in B$ and $x \in A$ or $x \in C$. Let's divide into cases:
1 - Suppose $x \in A$. In this case we know $ x \in A \cup (B \cap C)$
2 - Suppose $x \in B$ and $x \in C$. We know $x \in B \cap C$, by the definition of interessection. Hence, $x \in A \cup (B \cap C)$. So we always have $x \in A \cup (B \cap C)$. Hence, $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$, by the definition of subset.
It's almost good, but you jump over an important detail, which I print red.
If $x\in A$, then $x\in A\cup(B\cap C)$. (This is good).
If $\color{red}{x\notin A}$, then $x\in B$ (because $x\in A\cup B$) and $x\in C$ (because $x\in A\cup C$); therefore $x\in B\cap C$ and so $x\in A\cup(B\cap C)$.
As your proof is written you seem to be proving that if $x\in A\cup(B\cap C)$, then $x\in A\cup(B\cap C)$, which is obviously true, but is not the proof you have to write down.