Range of a Non-Constant Continuous Function

Detailed Solution

Since $\mathbb R$ is connected and $f$ is continuous, the image $S=f(\mathbb R)$ is connected, and the only connected subsets of $\mathbb R$ are intervals (possibly degenerate). Because $f$ is non-constant, $S$ contains at least two distinct values, so it is a non-degenerate interval — this rules out (D), since an interval with more than one point is automatically infinite, never finite. This proves the "interval" part of (A). To see $S$ need not equal $\mathbb R$, take $f(x)=\arctan x$, which is continuous and non-constant, yet $S=(-\pi/2,\pi/2)\neq\mathbb R$, ruling out (C). This same example also rules out (B): $S=(-\pi/2,\pi/2)$ happens to be open here, but $f(x)=\sin x$ gives $S=[-1,1]$, a closed interval, showing $S$ need not be open either. Thus the only property guaranteed in every case is that $S$ is a non-degenerate interval that need not be all of $\mathbb R$. The correct option is $\boxed{\mathrm{A}}$.