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 ${\beta }^{+}\mathrm{\notin }\gamma$ 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 $\xi \mathrm{\ne }\eta$ 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 ${\alpha }^{+}\mathrm{\ne }\xi$ 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.