If $U \subset S \subset X$, $U$ is open in $S$, and $S$ is open in $X$, then $U$ is open in $X$.

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

HINT: Work directly with closed sets. If $U$ is closed in $S$, there is an $F\subseteq X$ such that $U=F\cap S$, and $F$ is closed in $X$. But then $U$ is the intersection of two closed sets in $X$, so ...
If $x$ is an adherent point of $F$ in $X$, then $x$ is an adherent point of $S$ in $X$, hence $x\in S$. So $x$ is an adherent point of $F$ in $S$, hence $x\in F$.
$U$ open in $S$ which inherits the subspace topology as a subspace of $X$ so we have $\ U = M \cap S$ where $M$ is open in $X$. Now $U$ is the intersection of two open sets in $X \Rightarrow U$ is open.