Munkres Introduction to Topology: Section 17 Problem 4

If $A$ is closed in $X$, then we know that $A$ is equal to the compliment of some open set, $V$ in $X$. We write this as $A \ = \ X \ - \ V$. This means that $U \ - \ A \ = \ U \ - \ (X \ - \ V)$. By a basic set-theory identity, $A \ \cap \ B^{c} \ = \ A \ - \ B$, we get $U \ - \ (X \ - \ V) \ = \ U \ \cap \ (X \ - \ V)^{c} \ = \ U \ \cap \ V$. Obviously, this is an intersection of two open sets, and an intersection of two open sets will be open, so $U \ - \ A$ is open in $X$. On the other hand, we get $A \ - \ U \ = \ (X \ - \ V) \ - \ U \ = \ (X \ - \ V) \ \cap \ (X \ - \ U) \ \Rightarrow \ V^{c} \ \cap \ U^{c}$. An intersection of closed sets of closed, and the complement of an open set is closed, therefore this set is closed in $X$.