Every ordinal with its order topology has a basis of clopen sets, and is $T_1$, Hausdorff and regular

The set $T_\beta := \{\, \zeta \in \gamma : \beta < \zeta \,\}$ is open for every $\beta \in \gamma$: if $\beta < \zeta$ with $\zeta \in \gamma$ then $(\beta,\zeta]$ is a basic set with $\zeta \in (\beta,\zeta] \subseteq T_\beta$, by [A1] and transitivity in [L1].

The set $S_\xi := \{\, \zeta \in \gamma : \zeta < \xi \,\}$ is open for every $\xi \in \gamma$: if $\zeta < \xi$ then $[0,\zeta]$ is a basic set with $\zeta \in [0,\zeta] \subseteq S_\xi$, again by [A1] and transitivity.

Let $\xi \ne \eta$ in $\gamma$ and assume $\xi < \eta$ without loss of generality, by [L1]. Then $[0,\xi]$ and $(\xi,\eta]$ are basic open sets with $\xi \in [0,\xi]$, $\eta \in (\xi,\eta]$ and $[0,\xi] \cap (\xi,\eta] = \varnothing$ by [A1] and trichotomy; so $\gamma$ is Hausdorff, which is claim 3.

$\gamma \setminus [0,\beta] = T_\beta$ by trichotomy, so $[0,\beta]$ is closed by step 1.1 and [L2]; and $[0,\beta]$ is open, being basic.

$\gamma \setminus (\alpha,\beta] = S_{\alpha^{+}} \cup T_\beta$ by trichotomy, where $S_{\alpha^{+}} = [0,\alpha]$ is basic open and $T_\beta$ is open by step 1.1, so $(\alpha,\beta]$ is closed by [L2]; and it is open, being basic.

$\gamma \setminus \{\xi\} = S_\xi \cup T_\xi$ by trichotomy, which is open by steps 1.1 and 1.2 and [L2], so $\{\xi\}$ is closed.

Steps 2.1 and 2.2 exhaust $\mathcal{B}_\gamma$, so every basic set is clopen, which is claim 1.

Step 2.3 makes every singleton closed, so $\gamma$ is $T_1$ by [L3], which is claim 2.

Let $\xi \in \gamma$ and let $N$ be a neighbourhood of $\xi$; by [L4] there is a basic $B \in \mathcal{B}_\gamma$ with $\xi \in B \subseteq N$, and $B$ is closed by step 3.1 and open, hence a closed neighbourhood of $\xi$ inside $N$.

By step 4.1 every point of $\gamma$ has a neighbourhood base of closed neighbourhoods, so $\gamma$ is regular by [L5]; with step 3.2 it is $T_3$, which is claim 4.