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

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

In $(X,\mathcal{T}_{\mathrm{disc}})$ 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 $(X,\mathcal{T}_{\mathrm{disc}})$ let $C$ be closed and $x_0 \notin C$; the function $f$ with $f(x_0) = 1$ and $f(x) = 0$ for $x \ne x_0$ 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 $x_0 \notin C$, so the space is completely regular; taking $U := \{x_0\}$ and $V := C$ shows it is regular.

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

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

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

In $(Y,\mathcal{T}_{\mathrm{ind}})$ a separated pair $A, B$ has an empty member: if both were nonempty then $\overline{A} \cap B = Y \cap B = B \ne \varnothing$ 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 $(Y,\mathcal{T}_{\mathrm{ind}})$, given a separated pair with, say, $A = \varnothing$, the open sets $U := \varnothing$ and $V := Y$ separate them, and symmetrically when $B = \varnothing$; so the space is completely normal, and normal, disjoint closed sets being separated by [L2] and step 1.6.

In $(Y,\mathcal{T}_{\mathrm{ind}})$ both closed sets $\varnothing$ 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.