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Take distinct classes $p(x)\ne p(y)$. The points are not related by $R$, so by the defining negation there is a Hausdorff space $Z$ and a continuous map $f:X\to Z$ with $f(x)\ne f(y)$. Choose disjoint open neighborhoods $U$ and $V$ of those two values. Their inverse images $A=f^{-1}(U)$ and $C=f^{-1}(V)$ are disjoint open subsets of $X$ and are saturated by part (ii).

Hence $p^{-1}(p(A))=A$ and $p^{-1}(p(C))=C$, making $p(A)$ and $p(C)$ open in the quotient topology. They contain the chosen classes and are disjoint: an equivalence class meeting both would contradict saturation and $A\cap C=\varnothing$. Thus $X/R$ is Hausdorff. This quotient is the Hausdorff reflection by continuous maps; every continuous map from $X$ to a Hausdorff space factors uniquely through it.

Take distinct classes $p(x)\ne p(y)$. The failure of $x\mathcal{R}y$ supplies a Hausdorff space $Y$ and a continuous map $g:X\to Y$ with $g(x)\ne g(y)$. Choose disjoint open neighborhoods $V,W$ of these two images. Their inverse images $A=g^{-1}(V)$ and $B=g^{-1}(W)$ are disjoint open subsets of $X$, and part (ii) says both are saturated.

The quotient topology therefore makes $p(A)$ and $p(B)$ open, since their inverse images are precisely $A,B$. They are disjoint as well: an equivalence class meeting both would, by saturation, be contained in both. They contain $p(x)$ and $p(y)$ respectively. Thus $X/\mathcal{R}$ is Hausdorff. This construction is the Hausdorff reflection by continuous maps.