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

A subset is connected when its subspace topology admits no separation into two disjoint nonempty open sets. Equivalently, its only subsets that are both open and closed in that subspace are the empty set and the entire subset.

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 a continuous image of a connected space is connected.

The underlying set of the quotient space is the set of equivalence classes $X/R$. With quotient map $q(x)=[x]$, its quotient topology is defined by declaring $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. This makes $q$ a continuous surjection and is the finest topology with that property.