If $U$ is open in X, $A$ is closed in X, then $U-A$ is open in X, $A-U$ is closed in X. I want to make sure that the following proof is correct
$A$ is closed so we have that $X-A$ is open. But then $U\cap(X-A)$=$U\cap X-U\cap A$=$U-A\cap U=U\cap U-A\cap U=U\cap(U-A)$. Since $X-A$ and $U$ are open, then so must $U-A$ be open.
Im thinking about the solution for the second part.
Easier still, you have $U - A = U\cap A^c$, where $A^c = X - A$ is the complement of $A$. As $A$ is closed, $A^c$ is open, and the intersection of two opens is opens by the axioms of the topology. The other problem can be solved in a similar way.