The inverse of $R^{-1}$ is $R$. That is, $(R^{-1})^{-1}$ = $R$. I have to prove that this is true. In this case, R is a relation from the set A to the set B.
I know that $R = \{(x,y)\ \in \ A \times B \ | \ (x,y)\ \in \ R\}$. I also know that $R^{-1} = \{(y,x)\ \in \ B \times A \ | \ (x,y)\ \in \ R\}$. Since the two are supposed to be equal, I have to show that $R^{-1} \subseteq R$ and $R \subseteq R^{-1}$ to prove equality. I just don't know how to go about doing so.
You're not showing the two are equal. You're showing taking the inverse of the inverse relation gives you the original relation.
$$(R^{-1})^{-1}=\{(x,y)\in A \times B \mid (y,x) \in R^{-1}\}$$ $$ = \{(x,y)\in A \times B \mid (x,y) \in R\} \checkmark$$