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 $\min\{|x-y|,1\} = 0$ exactly when $|x-y| = 0$, that is $x = y$, by [L2]; for $i \ne j$ the value is $2 \ne 0$ 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 $\min\{|x-z|,1\} \le |x-z| \le |x-y|+|y-z|$ 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].

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

$N$ 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 $\min\{|x-y|,1\} < r$ holds exactly when $|x-y| < r$.

The saturation of $U \subseteq S$ is $U \cup \sigma[U \cap N]$: the class of $(x,i)$ is $\{(x,0),(x,1)\}$ for $x \ne 0$ 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 $\varepsilon, \delta > 0$ with $(-\varepsilon,\varepsilon)\times\{0\} \subseteq q^{-1}[V]$ 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 $t := \min\{\varepsilon,\delta\}/2$ one has $0 < t < \varepsilon$ and $t < \delta$ by [L2] and [L3], so $t \ne 0$, and $q(t,0) = q(t,1)$ by step 1.4 lies in $V \cap W$; hence $V \cap W \ne \varnothing$. 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.