= Solution
\b[No.] Take the <Hausdorff space> $X=\mathbb R$ and declare all nonzero points equivalent, leaving $0$ in its own class. The quotient has two points,
$$
\{[0],[\mathbb R\setminus\{0\}]\}.
$$
The nonzero class is open because its inverse image $\mathbb R\setminus\{0\}$ is open. However, the only saturated set containing $0$ but not the other class is $\{0\}$, which is not open. Thus every neighbourhood of $[0]$ is the whole quotient, so the two quotient points cannot have disjoint neighbourhoods. Therefore $\widetilde X$ need not be Hausdorff.
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