Continuous image of a real interval

A continuous function on any real interval has image a real interval. Given two image values and an intermediate value, choose preimages and apply the intermediate value theorem on the closed segment joining them, which stays inside the domain. Neither compactness nor boundedness of the original interval is needed. In particular $f(x)=\sin (1/x)/x$ on $(0,1]$ is a continuous surjection onto $\mathbb{R}$: its values are unbounded above and below, so its interval image contains every real number.