Past exam of the options I course of the University of Oxford / 2008 / E / E2

A subset is connected when it has no separation into two disjoint nonempty relatively open sets. It is path-connected when every pair of its points can be joined by a continuous path from $[0,1]$ lying in the subset.

Path-connectedness implies connectedness; the reverse fails in general, for example for the topologist's sine curve. They are equivalent for open subsets of every Euclidean space ${\mathbb{R}}^n$. More generally this holds for open subsets of locally path-connected spaces.

If $f(A)$ were separated by two disjoint nonempty relatively open sets, their preimages under the restriction $f{|}_A$ would be disjoint nonempty relatively open sets covering $A$. Continuity makes them open in $A$, contradicting its connectedness. Therefore $f(A)$ is connected.

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.