The preimage of continuous function on a closed set is closed.
Solution 1:
Yes, it looks right. Alternatively, given a continuous map $f: X \to Y$, if $D \subseteq Y$ is closed, then $X \setminus f^{-1}(D) = f^{-1}(Y \setminus D)$ is open, so $f^{-1}(D)$ is closed.
Solution 2:
Yes, your proof is correct, but since you are using sequences this works on metric spaces, not on topological ones.