= Solution
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>.
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