On an ordinal with its order topology the sets $[0,\beta]$ and $(\alpha,\beta]$ form a basis of clopen sets, the isolated points are exactly the non-limit ordinals, and the space is Hausdorff

For $\beta \in \gamma$ the set $[0,\beta]$ is open: if $\beta^{+} \in \gamma$ then $\xi \le \beta$ is equivalent to $\xi < \beta^{+}$ by [L5], so $[0,\beta] = \gamma_{<\beta^{+}}$, an open ray; and if $\beta^{+} \notin \gamma$ then $\beta^{+} \subseteq \gamma$, since every ordinal below $\beta^{+}$ is $\le \beta$ and so lies in $\gamma$, whence $\beta^{+} = \gamma$ by [L1] and $[0,\beta] = \gamma$, which is open.

For $\xi \ne \eta$ in $\gamma$ the trichotomy of [L1] gives $\xi < \eta$ after renaming. By [L5] the inequality $\xi < \eta$ gives $\xi^{+} \le \eta$, and since $\eta \in \gamma$, [L1] puts $\xi^{+}$ in $\gamma$ as well; so $\gamma_{<\xi^{+}}$ and $\gamma_{>\xi}$ are open rays, and they contain $\xi$ and $\eta$ respectively, since $\xi < \xi^{+}$ and $\xi < \eta$. They are disjoint: a common point $\zeta$ would satisfy $\zeta < \xi^{+}$, hence $\zeta \le \xi$ by [L5] and trichotomy, and $\xi < \zeta$ at once, which [L1] forbids. So claim 3 holds by [L7].

For $\alpha, \beta \in \gamma$ the set $(\alpha,\beta] = \gamma_{>\alpha} \cap [0,\beta]$ is open by [L4], being an intersection of two open sets.

Every set $[0,\beta]$ is closed, its complement being the open ray $\gamma_{>\beta}$.

Every set $(\alpha,\beta]$ is closed: its complement in $\gamma$ is $[0,\alpha] \cup \gamma_{>\beta}$, a union of an open set by step 1.1 and an open ray, hence open by [L4]. So every member of $\mathcal{B}_\gamma$ is clopen.

Claim 2, the isolated points. The point $0$ is isolated when $0 \in \gamma$, since $\{0\} = [0,0]$ is open by step 1.1; and a successor $\xi = \eta^{+} \in \gamma$ is isolated, since $\eta < \xi$ puts $\eta$ in $\gamma$ and $\{\xi\} = (\eta,\xi]$ is open by step 2.1.

$\mathcal{B}_\gamma$ is a basis. Let $U$ be open and $\xi \in U$; by [L2] and [L3] there is a set $B$ among $\gamma$, the open rays and the open intervals with $\xi \in B \subseteq U$. If $B = \gamma$ or $B = \gamma_{<b}$, then $[0,\xi]$ contains $\xi$ and lies inside $B$, since $\eta \le \xi$ gives $\eta < b$ in the second case. If $B = \gamma_{>a}$ or $B = (a,b)$, then $a < \xi$ and $(a,\xi]$ contains $\xi$ and lies inside $B$. In each case a member of $\mathcal{B}_\gamma$ sits between $\xi$ and $U$, and its members are open by step 1.1 and step 2.1, so [L3] applies and claim 1 is proved.

Conversely let $\xi \in \gamma$ be a limit ordinal. Were $\{\xi\}$ open, step 4.1 would supply $B \in \mathcal{B}_\gamma$ with $\xi \in B \subseteq \{\xi\}$, so $B = \{\xi\}$. If $B = [0,\beta]$ then $\beta = \xi$ and $0 \in B$, forcing $\xi = 0$, which no limit ordinal is. If $B = (\alpha,\xi]$ with $\alpha < \xi$ then $\alpha^{+} \le \xi$ by [L5], and $\alpha^{+} \ne \xi$ because $\xi$ is not a successor, so $\alpha < \alpha^{+} < \xi$ puts $\alpha^{+}$ in $B$ alongside $\xi$. Both cases are impossible, so $\{\xi\}$ is not open and $\xi$ is not isolated by [L6]; with step 3.2 this is claim 2.