Result R50: Set difference equals intersection with complement shows that \begin{equation} U \setminus F = U \cap (X \setminus F) \end{equation} The complement $X \setminus F$ is open in $T$ since $F$ is closed in $T$. A finite intersection of open sets is open, so that $U \cap (X \setminus F)$ and therefore $U \setminus F$ is open in $T$. $\square$