Continuous functions that map closed sets to closed sets

Detailed Solution

Option (C) is false as a general statement about all continuous functions: $f(x)=\arctan(x)$ is continuous, yet the closed set $\mathbb R$ maps to the non-closed set $(-\pi/2,\pi/2)$. For our specific $f$, which is assumed to map every closed set to a closed set, the key non-trivial fact (excluding the degenerate case where $f$ is constant, which trivially satisfies the hypothesis without being "proper") is that $f$ must be proper: if $f$ were not proper, there would be an unbounded closed set on which $f$ stays bounded, and a boundary value approached but never attained by $f$ on that set would break closedness of the image, e.g. $f(x)=1/(1+x^2)$ maps the closed set $\mathbb R$ to $(0,1]$, which is not closed. This is the standard characterization used to test option (B); options (A) and (D) each fail for the identity function $f(x)=x$, which is a non-constant, non-"monotone-and-surjective-only" style counterexample already covered by properness. Thus the intended correct option is $\boxed{\mathrm{B}}$.