Let $X$, $Y$, be metric spaces. Let $f: X \to Y$ be continuous. If $X$ is a compact metric space, show that $f^{-1}(K)$ is compact in $X$ whenever $K \subseteq Y$ is compact.
My proof is as follows:
Since $f$ is continuous and since $K$ is compact, $f^{-1}(K)$ is closed. Since $K$ is compact, $K \subseteq \bigcup\limits_{k=1}^{n}I_k$ where the $I_k$'s form a finite open cover of $K$. Again, since $f$ is continuous, $f^{-1}(\bigcup\limits_{k=1}^{n}I_k)=\bigcup\limits_{k=1}^{n}f^{-1}(I_k)$, which is a finite open cover of $f^{-1}(K)$. Thus $f^{-1}(K)$ is compact.