= Solution
\b[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 $0<r<\operatorname{dist}(x,\mathbb Z)$. The disjoint open sets
$$
U=(x-r,x+r),\qquad V=\mathbb R\setminus[x-r,x+r]
$$
are saturated; $U$ contains $x$ and $V$ contains all integers. Their images separate the two quotient points. These exhaust all pairs of distinct classes.
\b[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.
$$
\boxed{X_{\mathbb Z}\text{ is Hausdorff},\qquad X_{\mathbb Q}\text{ is not Hausdorff}.}
$$
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.
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