Suppose that $(X,\tau)$ is infinite and $T_0$, and suppose that it has no infinite $T_1$ subspace (which is what is needed for the five-space theorem).
intermediate motivational note. The five-space theorem (due to Sands and Ginsburg) says:
consider $\mathbb{N}=\{0,1,2\ldots,\}$ and these five topologies:
$\tau_0 = \{\emptyset, X\}$, the indiscrete topology,
$\tau_1 = \{\emptyset\} \cup \{\{n \ge p\}, p \in \mathbb{N}\}$, the final segment topology,
$\tau_2 = \{\mathbb{N}, \emptyset\} \cup \{\{n: n \le p\}: p \in \mathbb{N}\}$, the initial segment topology,
$\tau_3 = \{\emptyset\} \cup \{X \setminus F: F \subseteq \mathbb{N} \text{ finite }\}$, the cofinite topology,
$\tau_4 = \mathscr{P}(\mathbb{N})$, the discrete topology.
Then if $X$ is an infinite topological space, it contains a subspace homeomorphic to one of these 5 spaces. This is minimal because each of these 5 spaces has the property that any infinite subspace of it is homeomorphic to the whole space. The proof of this goes along these lines: first rule out an infinite indiscrete space, and get an infinite subspace that is $T_0$. (see intermediate note 2 below) Then the proof in this post shows that either we have an infinite $T_1$ space (from an antichain), or an initial or final segment subspace. Finally, for the infinite $T_1$ subspace we show that either there is an infinite cofinite subspace, or there is a countable discrete subspace, which I showed in this question. Hence, that question and this one are the most interesting parts of the proof of this 5-space theorem. Hence the added assumption that $X$ has no infinite $T_1$ subspace. So an infinite $T_0$ space contains one of the last 4 spaces, an infinite $T_1$ topology one of the last 2, as the first 3 are not $T_1$, and $T_1$ is hereditary. And an infinite $T_2$ space, only contains a countable discrete subspace for that reason..
end of inserted note
Define $x \le y$ iff $x \in \overline{\{y\}}$ iff $\forall O \in \tau: (x \in O) \to (y \in O)$, the so-called specialisation pre-order.
One checks that this is a pre-order: $x \le x$ for all $x$, is clear, and so is transitivity (consider the last reformulation which makes that clear).
For $T_0$-spaces it's a partial order: if $x \le y$ and $y \le x$, then $x = y$, for otherwise $x \neq y$ and the definition of $T_0$ says there is an open $O$ with $x \in O, y \notin O$ (which contradicts $x \le y$) or $x \notin O, y \in O$ (and this contradicts $y \le x$). So $x=y$ and we have a partial order.
inserted note 2
Note that $\le$ for general spaces $X$ only is a pre-order, i.e. reflexive and transitive, but not necessarily has the antisymmetric property that $x \le y \land y \le x \to x=y$; this property is in fact equivalent to the fact that $X$ is $T_0$. If we, in the proof of the five-spaces theorem, start with any infinite space $X$, we can consider this order, and define the standard equivalence relation $\sim$ induced by $\le$: $x \sim y$ iff $x \le y \land y \le x$, (note that being a preorder implies this is an equivalence relation) and note that we either have infinitely many different classes, or there is an infinite class of $\sim$. If the latter holds this class $A$ is a subspace with the indiscrete topology (any open set of it, if it contains one $x \in A$, it contains all of them) and we are done with the proof, or in the former case (infinitely many classes) pick one representative from each class and get an infinite $T_0$ subspace to continue the proof with.
end of inserted note 2
Now a standard fact in the theory of partial orders: any infinite partial order has a countable chain (a set of comparable elements) (up or down) or a countable antichain (a set of non-comparable elements). This follows e.g. from Ramsey's theorem and this and other proofs are here.
If an infinite antichain would exist, this would be an infinite $T_1$ subspace, as can easily be checked, so that options is ruled out (or we pass to that subspace and apply the cofinite vs discrete argument from before). So we have a countable descending chain $x_{n+1} \le x_{n}$ for all $n$ with maximum $x_0$ (all $x_n$ distinct in both cases), or a countable ascending chain $x_0$ as the minumum and $x_n \le x_{n+1}$ for all $n$.
In the last case, set $A = \{x_n :n =0,1,\ldots\}$ and suppose $O$ is open in $A$. Let $m = \min\{n: x_n \in O\}$, then $O \subseteq \{x_n: n \ge m\}$, but the reverse also holds (if $x_n \in A$, $n > m$ we have $x_m \le x_n$ and as $x_m \in O$, so also $x_n \in O$). Also, the sets $O_p:=\{x_n: n \ge p\}$ are all open in $A$ (this is clear for $p=0$ and if $p>0$ pick $O$ open with $x_p \in O, x_{p-1} \notin O$, as $x_{p} \not\le x_{p-1}$; we can do this to avoid all $x_l$ for $l < p$ (finite intersections) and then $p = \min\{n: x_n \in O\}$ and the above argument shows $O_p$ is indeed open. So open sets are essentially tails and all tails are open sets. So $x_n \to n$ is a homeomorphism of $A$ with $\mathbb{N}$ in the final segment topology.
If we have a descending chain, set $B =\{x_n, n=0,1,2\ldots\}$ and a similar argument shows that all open sets are initial segments and initial segments are open (use $\max$ instead of $\min$), so we have a homeomorphism of $B$ with $\mathbb{N}$ in the initial segment topology.
