 Solution

The unit circle is path-connected, since it is the continuous image of the interval $[0,2\pi ]$ under $\theta \mapsto (\cos \theta ,\sin \theta )$. It is therefore connected, and part (ii) makes $f(C)$ connected. A connected subset of $\mathbb{R}$ is an interval: omission of a point lying between two of its points would separate it by the two open rays about that point.

Moreover, the circle is compact, so the continuous image of a compact space is compact. Hence here the image is a closed bounded interval, possibly a singleton.