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 ${\mathrm{A}\mathrm{C}}_{\omega }$ 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 $E_n:=R\setminus (U_0\cup \cdots \cup U_n)$ 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 $\mathrm{\varnothing }$ 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 $x\mathrm{\notin }{R}_{<x}$. 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 $x_N\in [0_R,u]\subseteq U_{n_0}\cup \cdots \cup U_{n_p}\subseteq U_0\cup \cdots \cup U_N$, 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 ${\mathrm{A}\mathrm{C}}_{\omega }$ 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 $x<x^{\mathrm{\prime }}$ puts $x$ in $R_{<x^{\mathrm{\prime }}}$ 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.