My question: Theorem 22.3 (Smooth invariance of domain). Let $U \subset\mathbb{R}^n$ be an open subset, $S \subset\mathbb{R}^n$ an arbitrary subset, and $f : U \rightarrow S$ a diffeomorphism. Then $S$ is open in $\mathbb{R}^n$.

I can't understand why the set $S$ is not automatically open in $\mathbb{R}^n$. The mapping is a diffemorphism, which means it is continuous in both directions, so $S$ is open.

Answer: All you know a priori, is that open sets $V$ of $U$ satisfy: $f(V)$ is open in $S$, not that $f(V)$ is open in $\mathbb{R}^n$. So, $f(U)=S$ is open in $S$. The claim is then that $f(U)=S$ is actually open in $\mathbb{R}^n$, which is not the same thing and is not automatic. It requires proof.

This speaks of an important blind spot of open sets in topology, i.e. openness is relative. In particular, when considering some subset of a topological space, you should figure out if it is open in the subset, or in the ambient space.