A discrete space satisfies every axiom in the chain; an indiscrete space with two points is regular, completely regular, normal, completely normal and perfectly normal, and fails $T_0$

In $(X,{\mathcal{T}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}})$ every singleton is closed by [A1], so the space is $T_1$ by [L7], hence $T_0$.

In $(X,{\mathcal{T}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}})$ distinct points $x\mathrm{\ne }y$ have the disjoint open neighbourhoods $\{x\}$ and $\{y\}$, whose closures are themselves by [A1] and [L3]; so the space is Hausdorff and Urysohn.

In $(X,{\mathcal{T}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}})$ let $A$ and $B$ be separated, hence disjoint by [L2]; then $A$ and $B$ are themselves disjoint open sets containing them, so the space is completely normal, and in particular normal, every pair of disjoint closed sets being separated by [L2] and [L3].

In $(X,{\mathcal{T}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}})$ let $C$ be closed and $x_0\mathrm{\notin }C$; the function $f$ with $f(x_0)=1$ and $f(x)=0$ for $x\mathrm{\ne }{x}_0$ takes values in $[0,1]$ and is continuous by [L5], and it satisfies $f(x_0)=1$ and $f(y)=0$ for every $y\in C$, since $x_0\mathrm{\notin }C$, so the space is completely regular; taking $U:=\{x_0\}$ and $V:=C$ shows it is regular.

In $(Y,{\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}})$ the only open set containing $a$ is $Y$, which also contains $b$, and likewise with $a$ and $b$ exchanged; so no open set contains exactly one of them and the space is not $T_0$, hence not $T_1$, not Hausdorff and not Urysohn, and not metrizable by [L6].

In $(Y,{\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}})$ the closed sets are $\mathrm{\varnothing }$ and $Y$ by [A2], so $\bar{\mathrm{\varnothing }}=\mathrm{\varnothing }$ and $\bar{A}=Y$ for every nonempty $A$, the smallest closed superset of a nonempty set being $Y$.

In $(Y,{\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}})$ let $C$ be closed with $y_0\mathrm{\notin }C$; then $C\mathrm{\ne }Y$, so $C=\mathrm{\varnothing }$ by [A2], and the constant function $1$ is continuous by [L5] with $f(y_0)=1$ and $f[C]=\mathrm{\varnothing }\subseteq \{0\}$ vacuously, so the space is completely regular; and $U:=Y$, $V:=\mathrm{\varnothing }$ are disjoint open sets separating $y_0$ from $C$, so it is regular.

In $(X,{\mathcal{T}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}})$ every closed set is open by [A1], hence a $G_{\delta }$ by [L4]; with step 1.3 the space is perfectly normal.

In $(Y,{\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}})$ a separated pair $A,B$ has an empty member: if both were nonempty then $\bar{A}\cap B=Y\cap B=B\mathrm{\ne }\mathrm{\varnothing }$ by step 1.6, contradicting [L2].

By steps 1.1 to 1.5 the discrete space satisfies every axiom listed in claim 1, and the numbered forms follow, each numeral being its adjective together with $T_1$, which holds by step 1.1.

In $(Y,{\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}})$, given a separated pair with, say, $A=\mathrm{\varnothing }$, the open sets $U:=\mathrm{\varnothing }$ and $V:=Y$ separate them, and symmetrically when $B=\mathrm{\varnothing }$; so the space is completely normal, and normal, disjoint closed sets being separated by [L2] and step 1.6.

In $(Y,{\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}})$ both closed sets $\mathrm{\varnothing }$ and $Y$ are open by [A2], hence $G_{\delta }$ by [L4]; with step 3.2 the space is perfectly normal.

Steps 3.2, 4.1, 1.7 and 1.5 are claim 2, and step 3.1 is claim 1.