A space is $T_1$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology

(a) implies (b): fix $x \in X$ and let $y \in X \setminus \{x\}$; then $y \ne x$, so [A1] supplies an open $V$ with $y \in V$ and $x \notin V$, whence $y \in V \subseteq X \setminus \{x\}$.

(b) implies (c): let $F \subseteq X$ be finite and list it as $F = \{x_0, \dots, x_{n-1}\}$ by [L4], so that $F = \{x_0\} \cup \dots \cup \{x_{n-1}\}$; for $n = 0$ this reads $F = \varnothing$, which is closed by [L1].

(c) implies (d): let $U \in \mathcal{T}_{\mathrm{cof}}$; if $U = \varnothing$ then $U \in \mathcal{T}$ by [L1], and otherwise $X \setminus U$ is finite by [L3], hence closed by (c), hence $U$ is open.

(d) implies (a): let $x \ne y$ in $X$; the sets $X \setminus \{y\}$ and $X \setminus \{x\}$ have finite complements, so they lie in $\mathcal{T}_{\mathrm{cof}}$ by [L3] and hence in $\mathcal{T}$ by (d), and they witness the $T_1$ condition, since $x \in X \setminus \{y\}$, $y \notin X \setminus \{y\}$, $y \in X \setminus \{x\}$ and $x \notin X \setminus \{x\}$.

By step 1.1 the set $X \setminus \{x\}$ is a neighbourhood of each of its points, hence open by [L2], so $\{x\}$ is closed by [L1]; this completes the implication (a) implies (b).

By step 1.2 and (b) the set $F$ is a union of $n$ closed sets, hence closed by [L1]; this completes the implication (b) implies (c).

The four implications of steps 2.1, 2.2, 1.3 and 1.4 close the cycle (a) implies (b) implies (c) implies (d) implies (a), so the four conditions are equivalent.

In particular $\mathcal{T}_{\mathrm{cof}}$ itself satisfies (d) with $\mathcal{T} = \mathcal{T}_{\mathrm{cof}}$, so the cofinite topology on any set is $T_1$ by step 3.1, and by (d) it is contained in every $T_1$ topology on that set; this is the final assertion of the statement.