Past exam of the topology course of the University of Oxford / 2023 / 2 / a / iv

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is finite and cannot be all of $\mathbb{R}$.

Suppose the quotient space had distinct points with disjoint open neighbourhoods $U,V$. Since the quotient map $p$ is surjective, both preimages $p^{-1}(U),p^{-1}(V)$ would be nonempty. Since $p$ is continuous, these would be disjoint open subsets of $\mathbb{R}$ with its cofinite topology, contradicting the observation above. Thus any such quotient space with at least two points is not Hausdorff.