Solution: The main thing that we will be using is if $A,B\subseteq X$, then $A\setminus B = A\cap B^c$, where $B^c = X\setminus B$. Note that \[ U\setminus F = U \cap F^c. \] Since, $F$ is closed so its complement will be open and union of open sets is open, therefore, $U\setminus F$ is open. Similarly, \[ F\setminus U = F \cap U^c. \] As $U$ is open so $U^c$ is closed and intersection of closed sets is closed, so $F\setminus U$ is closed.