Every successor ordinal is compact in its order topology and every limit ordinal is not; and, assuming countable choice, $\omega_1$ is countably compact and sequentially compact while $\omega_1 + 1$ is compact

For claim 1 let $\gamma := \delta^{+}$, so that $\delta$ is the greatest element of $\gamma$ and $[0,\delta] = \gamma$; let $\mathcal{U}$ be an open cover of $\gamma$ and put $S := \{\, \eta \in \gamma : \text{finitely many members of } \mathcal{U} \text{ cover } [0,\eta] \,\}$.

For claim 2 let $\lambda$ be a limit ordinal; the family $\{\, [0,\beta] : \beta \in \lambda \,\}$ consists of open sets by [L2] and covers $\lambda$, since $\xi \in [0,\xi]$ for every $\xi \in \lambda$.

For claim 3 assume $\mathrm{AC}_\omega$ and let $(y_k)$ be a sequence in $\omega_1$; its range is at most countable by [L7], so [L6] gives $\sigma := \sup\{\, y_k : k \in \mathbb{N} \,\} \in \omega_1$, and the set $\{\, k \in \mathbb{N} : y_k \le \sigma \,\}$ is all of $\mathbb{N}$ and in particular infinite.

Let $\eta \in \gamma$ and suppose $[0,\zeta]$ is covered by finitely many members of $\mathcal{U}$ for every $\zeta < \eta$. Some $U \in \mathcal{U}$ contains $\eta$, and [L2] gives $B \in \mathcal{B}_\gamma$ with $\eta \in B \subseteq U$. If $B = [0,\beta]$ then $\eta \le \beta$ and $[0,\eta] \subseteq [0,\beta] \subseteq U$, so $\{U\}$ covers $[0,\eta]$. If $B = (\alpha,\beta]$ then $\alpha < \eta \le \beta$, and $[0,\eta] \subseteq [0,\alpha] \cup (\alpha,\eta] \subseteq [0,\alpha] \cup U$ by [L3], so a finite cover of $[0,\alpha]$ with $U$ adjoined covers $[0,\eta]$. Either way $\eta \in S$.

A finite subfamily of the cover of step 1.2 is empty, and then covers only $\varnothing \ne \lambda$, or is $[0,\beta_0], \dots, [0,\beta_n]$ with union $[0,\beta]$ for $\beta$ the greatest of the $\beta_j$, which exists by [L3]; and $\beta^{+} \in \lambda$ by [L3] while $\beta^{+} \notin [0,\beta]$. So no finite subfamily covers $\lambda$ and $\lambda$ is not compact, which is claim 2.

By [L5] and [L6] the set $\{\, \xi \in \omega_1 : \{k : y_k \le \xi\} \text{ is infinite} \,\}$ is a nonempty set of ordinals, $\sigma$ belonging to it by step 1.3, so it has a least element $\tau$ by [L3]; then $P := \{\, k \in \mathbb{N} : y_k \le \tau \,\}$ is infinite while $\{\, k : y_k \le \alpha \,\}$ is finite for every $\alpha < \tau$.

By [L4] applied to the well-ordered $\gamma$, step 2.1 gives $S = \gamma$; in particular $\delta \in S$, so finitely many members of $\mathcal{U}$ cover $[0,\delta] = \gamma$. As $\mathcal{U}$ was arbitrary, $\gamma = \delta^{+}$ is compact, which is claim 1.

Let $i \mapsto m_i$ be the strictly increasing enumeration of $P$ given by [L8]; then $(y_{m_i})$ is a subsequence of $(y_k)$ by [L9], and every one of its terms satisfies $y_{m_i} \le \tau$.

$y_{m_i} \to \tau$. Let $U$ be open with $\tau \in U$ and take $B \in \mathcal{B}_{\omega_1}$ with $\tau \in B \subseteq U$ by [L2]. If $B = [0,\beta]$ then $\tau \le \beta$ and every term satisfies $y_{m_i} \le \tau \le \beta$, so all terms lie in $B$. If $B = (\alpha,\beta]$ then $\alpha < \tau \le \beta$, the set $\{k : y_k \le \alpha\}$ is finite by step 2.3, so $\{\, i : y_{m_i} \le \alpha \,\}$ is finite, the map $i \mapsto m_i$ being injective; hence $\alpha < y_{m_i} \le \tau \le \beta$ for all large $i$ and the terms lie in $B$ from some index on. So $\omega_1$ is sequentially compact.

By [L10] the space $\omega_1$ is therefore countably compact; it is not compact by step 2.2, being a limit ordinal by [L5]; and $\omega_1 + 1 = \omega_1^{+}$ is compact by step 3.1 and [L11]. This is claim 3, and with claims 1 and 2 at steps 3.1 and 2.2 the theorem is proved.