 Solution

The circle is path-connected, using circular arcs, and hence connected. Its continuous image is connected by part (ii). A connected subset of the real line is an interval: if it contained values on opposite sides of a missing value $c$, intersection with $(-\infty ,c)$ and $(c,\infty )$ would separate it. Thus $f(C)$ is an interval. It is also compact, so it is a closed bounded interval, possibly reduced to one point.