 Solution

Collapsing the integers gives a Hausdorff space. For two distinct noninteger points, choose disjoint small open intervals around them that avoid $\mathbb{Z}$. Each interval is saturated under the saturation of a subset operation, so its image is an open neighbourhood of its class. For a noninteger $x$ and the collapsed class, choose $0<r<\mathrm{dist}(x,\mathbb{Z})$. The disjoint open setsare saturated; $U$ contains $x$ and $V$ contains all integers. Their images separate the two quotient points. These exhaust all pairs of distinct classes.

Collapsing the rationals does not give a Hausdorff space. The preimage of the singleton collapsed point is $\mathbb{Q}$, which is not closed in $\mathbb{R}$. Therefore that quotient singleton is not closed, whereas every singleton of a Hausdorff space is closed.Equivalently, every nonempty saturated open subset of $\mathbb{R}$ meets the dense set $\mathbb{Q}$ and therefore contains it all. Every nonempty quotient open set contains the collapsed point, making separation from any other quotient point impossible.