Every closed initial segment of the long ray is compact; the long ray is not compact; and, assuming countable choice, it is countably compact and not Lindel"of

For claim 1 fix $u \in R$ and a family $\mathcal{U}$ of open subsets of $R$ with $[0_R,u] \subseteq \bigcup \mathcal{U}$, and put $D := \{\, y \in [0_R,u] : \text{finitely many members of } \mathcal{U} \text{ cover } [0_R,y] \,\}$. Then $0_R \in D$, since $0_R$ lies in some member of $\mathcal{U}$ and $[0_R, 0_R] = \{0_R\}$, and $u$ is an upper bound of $D$; so [L2] gives $s := \sup D$, and $s \le u$ because $u$ is an upper bound.

For claim 3 the family $\mathcal{R} := \{\, R_{<x} : x \in R \,\}$ is an open cover of $R$: its members are open by [L1], and every $y \in R$ lies in $R_{<x}$ for some $x$, since $R$ has no greatest element.

For claim 2 assume $\mathrm{AC}_\omega$ and let $\mathcal{U}$ be an at most countable open cover of $R$ with no finite subcover; it is nonempty, so [L6] lists it as $(U_n)_{n \in \mathbb{N}}$, and $E_n := R \setminus (U_0 \cup \dots \cup U_n)$ is nonempty for every $n$, so countable choice supplies a sequence $(x_n)$ with $x_n \in E_n$. The range $\{\, x_n : n \in \mathbb{N} \,\}$ is at most countable by [L6], so [L3] gives an upper bound $u \in R$ for it.

$s$ lies in some $U \in \mathcal{U}$, and [L1] gives a basic $B$ with $s \in B \subseteq U$, where $B$ is $R$, an open ray or an open interval. If $B$ is $R$ or a lower ray $R_{<b}$, then every $z \le s$ satisfies $z < b$, so $[0_R,s] \subseteq B \subseteq U$ and $\{U\}$ covers $[0_R,s]$. Otherwise $B$ is $R_{>a}$ or $(a,b)$ with $a < s$, and $a$ is not an upper bound of $D$, so some $y \in D$ has $a < y \le s$; a finite $\mathcal{F} \subseteq \mathcal{U}$ covers $[0_R,y]$, and $[0_R,s] \subseteq [0_R,y] \cup (a,s] \subseteq \bigcup \mathcal{F} \cup B$, so $\mathcal{F} \cup \{U\}$ covers $[0_R,s]$. In every case $s \in D$.

No finite subfamily of $\mathcal{R}$ covers $R$: the empty subfamily covers $\varnothing$ and $R$ is nonempty, while a subfamily $R_{<x_0}, \dots, R_{<x_n}$ has union $R_{<x}$ for $x$ the greatest of the $x_j$, which exists because the order is linear and the list finite, and $x \notin R_{<x}$. So $R$ is not compact, which is claim 3.

$s = u$. Suppose $s < u$, and keep $B$ and $U$ from step 2.1, together with a finite $\mathcal{G} \subseteq \mathcal{U}$ covering $[0_R,s]$, which step 2.1 provides. If $B$ is $R$ or $R_{>a}$, then $(s,u] \subseteq B \subseteq U$ and $\mathcal{G} \cup \{U\}$ covers $[0_R,u]$, putting $u$ in $D$ and forcing $u \le s$, contrary to $s < u$. If $B$ is $R_{<b}$ or $(a,b)$, then $s < b$ and $s < u$, so the lesser of $b$ and $u$ is strictly above $s$ and [L2] gives $z$ with $s < z$ and $z$ below that lesser element; then $z \le u$ and $(s,z] \subseteq B$, so $\mathcal{G} \cup \{U\}$ covers $[0_R,z]$ and $z \in D$ with $z > s$, contradicting $s = \sup D$.

By steps 2.1 and 3.1 the element $u$ lies in $D$, so finitely many members of $\mathcal{U}$ cover $[0_R,u]$; as $\mathcal{U}$ was arbitrary, [L5] makes $[0_R,u]$ a compact subset of $R$, which is claim 1.

By claim 1 the set $[0_R,u]$ is covered by finitely many of the $U_n$, say by $U_{n_0}, \dots, U_{n_p}$; let $N$ be the greatest of $n_0, \dots, n_p$. Then $x_N \le u$, so $x_N \in [0_R,u] \subseteq U_{n_0} \cup \dots \cup U_{n_p} \subseteq U_0 \cup \dots \cup U_N$, contradicting $x_N \in E_N$. So no such $\mathcal{U}$ exists and $R$ is countably compact, which is claim 2.

For claim 4 assume $\mathrm{AC}_\omega$ and let $\mathcal{V} \subseteq \mathcal{R}$ be an at most countable subfamily of the cover of step 1.2. The map $x \mapsto R_{<x}$ is injective, since $x < x'$ puts $x$ in $R_{<x'}$ and not in $R_{<x}$, so $A := \{\, x \in R : R_{<x} \in \mathcal{V} \,\}$ is at most countable and [L3] gives it an upper bound $w \in R$; then $R_{<x} \subseteq R_{<w}$ for every $x \in A$, so $\bigcup \mathcal{V} \subseteq R_{<w}$ and $w$ is covered by no member of $\mathcal{V}$. So $\mathcal{R}$ has no at most countable subcover and $R$ is not Lindelöf, which is claim 4; with claims 1, 2 and 3 at steps 4.1, 5.1 and 2.2 the theorem is proved.