My book in Analysis says:
Let $(S,d)$ be a metric space. $S$ is open in $S$ and the empty set $\emptyset$ is open in $S$.
Fair enough. But then they proceed...
A subset $E$ of $S$ is closed if its complement $S\setminus E$ is an open set. In other words, $E$ is closed if $E=S\setminus U$ where $U$ is an open set.
How is this not a contradiction? We have that $S=S\setminus\emptyset$. Therefore, $S$ should be closed. But $S$ is also open in $S$.