Keep in mind the proof is for a first class in probability theory!
\begin{align*} P(A \cap (B \cup C^C)) &= P((A \cap B) \cup (A \cap C^c)) \\ &= P(A \cap B) + P(A \cap C^c) - P(A \cap B \cap C^c) \\ &= P(A)(P(B) + P(C^c)) - P(A)P(B|A)P(C^c|B \cap A)) \\ &= P(A)(P(B) + P(C^c)) - P(A)P(B)P(C^c) \\ &= P(A) (P(B) + P(C^c) -P(B)P(C^c)) \\ &= P(A)(P(B \cup C^c)) \end{align*}
so they are independent.