Suppose $f(X)=U\cup V$ were a separation into two disjoint nonempty relatively <open sets>. <Continuity> makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$, and they are disjoint, nonempty and cover $X$. This contradicts <connectedness> of $X$. Hence $\boxed{f(X)\text{ is connected}}$.

Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / E / E4 / a / ii / Solution

 Solution (source code)