$f$ is continuous if and only if, $f(x_d) \to f(x)$ where $(x_d)_{d \in D}$ is a net such that $x_d \to x$
Solution 1:
The very last part is better stated as: as $f(x) = y \in U$, by definition $x \in f^{-1}[U]= V$. So $V$ is closed under limits of nets and hence closed by the lemma.
"Psycho-notionally" I'd use $C$ or $F$ for a closed subset, personally I associate the letters $U$,$V$ and $O$ with open sets. It's minor, but proofs should minimise possible confusion, IMHO. And then I'd use, say, $D=f^{-1}[C]$ next for the inverse image we want to be closed (the next letter). It's easier on my memory.