I'm just trying to verify that if two sets are independent, then the complement of one set is still independent of the other set.
Therefore this logic should work: If $A$ and $B$ are independent then
$P(A\cap B) = P(A)P(B)$
and then $P(A^c \cap B) = P(A^c)P(B)$.
Your way is right. Really, this proposition works for $n$ arbitrary: if $A_1,...,A_n$ are independent, $\phi$ a permutation of $\{1,...,n\}$ and $1\leq m \leq n$, then $A_{\phi(1)}^c, ..., A_{\phi(m)}^c, A_{\phi(m+1)}, ... ,A_{\phi(n)}$ are independent. The proof of this is by induction on $m$, you proved there case base
