A separation of $f(X)$ would pull back under the <continuous> map to two disjoint nonempty <open sets> covering $X$: nonemptiness follows because their points belong to the image. This contradicts <connectedness> of $X$. Hence \b[a <continuous> image of a <connected> space is <connected>].

Solution

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

 Solution (source code)