Exercise 2.2.4 (Openness in the original and relative topologies)

Proof. If $S$ is open in $X$, then every open subset of $S$ will be open in $X$ as well since it is an intersection of an open set in $X$ with $S$, which is as well open. Conversely, if every relatively open set in $S$ is open in $X$, then, in particular, $S$, which is trivially open in $S$, must be open in $X$ as well. □

Proof. Suppose that $S$ is closed in $X$. Let $C$ be an arbitrary relatively closed set in $S$. Since $S\setminus C$ is open in $X$, we can find an open set $U$ in $X$ such that $U\cap S=S\setminus C$. We thus obtain

Since $X\setminus S$ and $U$ are both open in $X$, $X\setminus C$ must be open as well and thus, $C$ must be closed in $X$. Conversely, if every relatively closed set in $S$ is also closed in $X$, then, in particular, $S$, which is trivially closed in $S$, must be closed in $X$ as well. □