Just to elaborate on your observation:
You could proceed from the following step too, but that will be a rather unnecessarily complicated way of doing. Look at the following:
$P(A^c \cap B)=P(A^c)-P(A^c\cap B^c)$
Now use the De-Morgan's Law: $A^c \cap B^c=(A \cup B)^c$
So, what does that mean, $\begin{align*}P((A \cup B)^c)&=1-P(A \cup B)\\&=1-P(A)-P(B)+P(A \cap B)\\&\overset{ind}{=}1-P(A)-P(B)+P(A) \cdot P(B)\\&=(1-P(A))\cdot(1-P(B))\\&=P(A^c)\cdot P(B^c)\end{align*}$
Now, you have proved on the way that $A^c$ and $B^c$ are independent iff (try to prove one of the direction I haven't proved!) $A$ and $B$ are independent.
Now to complete, put this in there, $\begin{align*}P(A^c \cap B)&=P(A^c)-P(A^c) \cdot P(B^c)\\&=P(A^c)[1-P(B^c)]\\&=P(A^c) \cdot P(B)\end{align*}$
Hope you're convinced!