I tried to solve the following.
Let $f:X_1\rightarrow X_2$ be a bijective map with inverse $g$. Show that for any subset $U$ of $X_1$, we have that $f(U)=g^{-1}(U)$.
$h\in g^{-1}(U)$ means that $h\in X_2$ and so due to bijectivity, there exists a unique $a\in X_1$ such that $f(a)=h \in g^{-1}(U)$, so $g(f(a))\in U$. As $g$ is the inverse of $f$, we have that $a\in U$ and therefore $f(a)\in U$.
Suppose $b\in f(U)$, that means that there exists a unique $c \in U$ such that $f(c)=b$. So $b\in f(U)\subseteq X_2$, $g(b)=c$, by definition of inverse. Hence $b\in g^{-1}(U)$.
Is the proof correct? May I have some feedback, please?
This is correct, but it also follows immediately from $f = g^{-1}$.