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

Let $C$ be any equivalence class. Since $A$ is both open and closed, $C\cap A$ is both open and closed in the subspace topology on $C$. Its connectedness makes $C\cap A$ either empty or all of $C$. Thus an equivalence class meeting $A$ lies wholly in $A$. Hence $A$ is saturated, a union of whole equivalence classes.

If $X$ is connected, its continuous quotient image is connected by part (a). Conversely, suppose $X$ is disconnected. It has a nonempty proper subset $A$ both open and closed. Part (i) makes $A$ and its complement saturated, so $q^{-1}(q(A))=A$ and $q^{-1}(q(X\setminus A))=X\setminus A$. The quotient topology therefore makes these disjoint nonempty images open, and they cover $X/R$. Thus the quotient is disconnected. This proves the two spaces are connected simultaneously.