Two copies of $\mathbb{R}$ glued along $\mathbb{R}\setminus \{0\}$ give a non-Hausdorff quotient of a metrizable space, by an open quotient map

$d$ satisfies (M1) and (M2): for $i=j$ one has $\min \{\mathrm{\mid }x-y\mathrm{\mid },1\}=0$ exactly when $\mathrm{\mid }x-y\mathrm{\mid }=0$, that is $x=y$, by [L2]; for $i\mathrm{\ne }j$ the value is $2\mathrm{\ne }0$ and the two points differ; and both clauses are symmetric in the two arguments.

$d$ satisfies (M3). If the two outer points share a tag, the left side is at most $1$; a middle point with the same tag gives $\min \{\mathrm{\mid }x-z\mathrm{\mid },1\}\le \mathrm{\mid }x-z\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }y-z\mathrm{\mid }$ when both right-hand terms are below $1$, and a right-hand side of at least $1$ otherwise, while a middle point with the other tag gives a right-hand side of $4$. If the two outer points have different tags, the left side is $2$ and the middle point shares a tag with at most one of them, so at least one right-hand term is $2$.

$\sigma$ is a homeomorphism of $S$: it is its own inverse, and it carries a set $U$ with traces $U_0,U_1$ to the set with traces $U_1,U_0$, so it preserves openness by [A1].

$a\mathrm{\ne }b$, the classes $\{(0,0)\}$ and $\{(0,1)\}$ being distinct; and $q(t,0)=q(t,1)$ for every $t\mathrm{\ne }0$.

$N$ is open in $S$, both of its traces being $\mathbb{R}\setminus \{0\}$, which is open by [L3].

For $0<r\le 1$: $B_d((x,i),r)=(x-r,x+r)\times \{i\}$, since a point with the other tag is at distance $2\ge r$, and for the same tag $\min \{\mathrm{\mid }x-y\mathrm{\mid },1\}<r$ holds exactly when $\mathrm{\mid }x-y\mathrm{\mid }<r$.

The saturation of $U\subseteq S$ is $U\cup \sigma [U\cap N]$: the class of $(x,i)$ is $\{(x,0),(x,1)\}$ for $x\mathrm{\ne }0$ and $\{(x,i)\}$ for $x=0$, so saturating adds exactly the swapped copies of the points of $U$ off the two origins.

Suppose $V,W$ are open in $L$ with $a\in V$, $b\in W$. Then $q^{-1}[V]$ and $q^{-1}[W]$ are open in $S$ by [A2], containing $(0,0)$ and $(0,1)$ respectively, so by [A1] and [L3] there are $\epsilon ,\delta >0$ with $(-\epsilon ,\epsilon )\times \{0\}\subseteq q^{-1}[V]$ and $(-\delta ,\delta )\times \{1\}\subseteq q^{-1}[W]$.

${\mathcal{T}}_d$ is the disjoint union topology. If $W$ is $d$-open and $(x,i)\in W$, a ball of radius $r\le 1$ inside $W$ is $(x-r,x+r)\times \{i\}$ by step 2.1, so each trace $W_i$ is open by [L3], whence $W$ is open in $S$ by [A1]; conversely if $U$ is open in $S$ and $(x,i)\in U$, then $U_i$ is open, so [L3] gives $\rho >0$ with $(x-\rho ,x+\rho )\subseteq U_i$, and the ball of radius $\min \{\rho ,1\}$ lies in $U$ by step 2.1 and [L2].

By steps 1.3, 1.5 and 2.2 the saturation of an open $U$ is the union of the open sets $U$ and $\sigma [U\cap N]$, hence open; so $q[U]$ is open in $L$ by [A2], and $q$ is an open quotient map by [L4]. This is claim 2.

With $t:=\min \{\epsilon ,\delta \}\mathrm{/}2$ one has $0<t<\epsilon$ and $t<\delta$ by [L2] and [L3], so $t\mathrm{\ne }0$, and $q(t,0)=q(t,1)$ by step 1.4 lies in $V\cap W$; hence $V\cap W\mathrm{\ne }\mathrm{\varnothing }$. As $V$ and $W$ were arbitrary, $a$ and $b$ have no disjoint open neighbourhoods, and $L$ is not Hausdorff by [A3]. This is claim 3.

By steps 1.1, 1.2 and 3.1 the function $d$ is a metric inducing the topology of $S$, so $S$ is metrizable and hence Hausdorff by [A3]. This is claim 1.

By step 4.1 the space $S$ is Hausdorff, by step 3.2 the map $q$ is a quotient map, and by step 3.3 the quotient $L$ is not Hausdorff; so a quotient of a Hausdorff space need not be Hausdorff, which refutes the claim.