Suppose $f$ is a continuous real-valued function on Euclidean space $\bf{R}^n$ with the property that there is a positive number $c$ such that $|f(x)| \ge c \cdot ||x||$ for all $x \in \bf{R}^n$. Show that if $K$ is a compact set of real numbers, then its inverse image under $f$, $f^{-1}(K)$, also is compact.
I attempted to use the definition of sequential compactness to arrive at the result, but in my proof realized I never used the condition that $|f(x)| \ge c \cdot ||x||$. Is there an error in my proof? Or is this statement true in general for continuous functions?
Proof. Let $K \subset\bf{R}$ be compact and $f: \bf{R}^n \to \bf{R}$. Consider $\{x_n\} \subset f^{-1}(K)$. Then, by continuity of $f$, $\{x_n\}$ has a convergence subsequence with limit in $f^{-1}(K)$ iff $\{f(x_n)\}$ has a convergent subsequence with limit $y\in K$. By compactness of $K$, and hence sequential compactness of $K$, this is the case. $\square$
Suppose that $K$ is compact, it is closed, since $f$ is continuous, $f^{-1}(K)$ is also closed.
Since $K$ is compact, there exists $M>0$ such that for every $y\in K, |y|<M$. Let $x\in f^{-1}(K), M\geq |f(x)|\geq c\|x\|$ implies that $\|x\|\leq {M\over c}$ and $f^{-1}(K)$ is bounded and closed henceforth compact.
For your proof, you assert that $x_n\in f^{-1}(K)$ has a convergent subsequence by continuity of $f$, and this is not true.