Let $X$ be a set with subsets $A$ and $B$. Prove
$$X\setminus(A\setminus B) = B\cup (X\setminus A).$$
I know that in order to prove equality, I must show the left side is a subset of the right, and the right is a subset of the left. Do I let an arbitrary element belong to the left side and show it is also in the right side? And vice versa?
Yes, that is generally how you would go about such a problem (a so-called "element-chasing proof"), but I would encourage you to go about these problems in an algebraic way whenever possible (it's much more efficient, as the linked to example shows): \begin{align*} B\cup(X\setminus A) &=B\cup(X\cap A^c) & \text{(by definition)}\\[0.5em] &= (B\cup X)\cap(B\cup A^c) & \text{(distributivity)}\\[0.5em] &= X\cap(B\cup A^c) & \text{(since $B\subseteq X$)}\\[0.5em] &= X\cap(A\cap B^c)^c & \text{(DeMorgan)}\\[0.5em] &=X\setminus(A\setminus B) \end{align*}