Suppose $S$ is a subspace of the topological space $X$. If $U \subset S \subset X$, $U$ is open in $S$, and $S$ is open in $X$, then $U$ is open in $X$. Same is true with "closed" instead of "open".
I've solved the open version, but for closed version I get a this expression for $X\setminus U=(S\cup V_1) \setminus (S\setminus V_2)$, which im trying to show is open in $X$, which looks hopeless. $V_1$ and $V_2$ are open in $X$.
This is exercise 3.5 of Lee topological manifolds