$\omega + 1$ as a convergent sequence together with its limit, and, assuming countable choice, $[0, \omega_1)$, in which every sequence lies inside an at most countable initial segment

Give every ordinal its order topology (The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis), under which it is $T_3$ — that is $T_1$, Hausdorff and regular (Every ordinal with its order topology has a basis of clopen sets, and is $T_1$, Hausdorff and regular). Two ordinals are worked here.

The space $\omega + 1$. By the successor clause of ordinal addition (Ordinal addition $\alpha + \beta$), $\omega + 1 = \omega^{+} = \omega \cup \{\omega\}$, so the space is the set of natural numbers together with one extra point on top ($\omega$ is the least limit ordinal). Then:

1. Every $n \in \omega$ is isolated: $\{0\} = [0,0]$ and $\{m^{+}\} = (m, m^{+}]$ are basic open sets.
2. The sequence $x_k := k$ ($k \in \mathbb{N}$) converges to $\omega$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), and it converges to no other point of $\omega+1$.

So $\omega + 1$ is, as a topological space, exactly a convergent sequence together with its limit, and $\omega$ is its unique non-isolated point.

The space $[0,\omega_1) = \omega_1$. Let $\omega_1$ be the first uncountable ordinal (The first uncountable ordinal $\omega_1 := \aleph(\omega)$), so that $\omega_1$ is a limit ordinal, every ordinal below it is at most countable, and $\omega_1$ itself is uncountable ($\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF). As a set, $[0,\omega_1)$ is $\omega_1$, an ordinal being the set of ordinals below it (Ordinal (von Neumann)). Assuming the Axiom of Countable Choice (The Axiom of Countable Choice ($\mathrm{AC}_\omega$)):

3. Every sequence $(x_k)$ in $\omega_1$ has an at most countable range, so there is $\alpha < \omega_1$ with $x_k \le \alpha$ for all $k$; hence the whole sequence lies inside the initial segment $[0,\alpha] = \alpha^{+}$, which is an ordinal below $\omega_1$ and is at most countable.
4. Consequently no sequence in $\omega_1$ has a range cofinal in $\omega_1$ (Cofinal subset of an ordinal).

Clause 3 is the fact the deleted Tychonoff plank consumes, and it is the reason $[0,\omega_1)$ behaves unlike any metrizable space: a sequence can never approach the "top" of $\omega_1$, because there is no top to approach along a sequence.

• $\omega + 1$ is the smallest interesting ordinal space. Every ordinal $\gamma \le \omega$ is discrete (The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis), so $\omega+1$ is where a non-isolated point appears for the first time, and it appears as the limit of the obvious sequence.

• Clause 3 is where the countable choice enters and where it stays. It is inherited from Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$, so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable and from nothing else on this page; clauses 1, 2 and the $T_3$ property of both spaces are theorems of ZF (Every ordinal with its order topology has a basis of clopen sets, and is $T_1$, Hausdorff and regular).

• What clause 4 rules out. No sequence in $[0,\omega_1)$ can be used to approximate the space from below, which is why arguments about $\omega_1$ are written with arbitrary at most countable sets rather than with sequences, and why the plank argument on this page bounds a set of ordinals rather than taking a limit of a sequence.