While solving an exercise in Munkres' "Topology" (Ex. 12(c) Sect. 24) I have tried proving the following lemma:
Let us denote the minimal uncountable well-ordered set by $S_{\Omega}$. Then $S_{\Omega}$ with the order topology satisfies the first countability axiom, i.e. for every point of $S_{\Omega}$ there exists a countable basis.
My Attempt
Let $a\in S_{\Omega}$ be arbitrary and let $a_0 = \min S_{\Omega}$. Since $S_{\Omega}$ is well-ordered, $a$ has an immediate successor, call it $b$. Now, by the definition of $S_{\Omega}$, the section by $a$ (that is, $[a_0, a[$) is countable, hence we can write it as $\{a_i\}_{i \in I}$ where $I$ is finite. Then it is easy to verify that $\{U_i = ]a_i, b[\}_{i \in I} \cup \{[a_0, b[\}$ is a countable basis for $a$.
My question
I would like to know if the claim is true and if my proof is good enough to show its truth because I couldn't find anything on it.
As always, any comment or answer is much appreciated and let me know if I can explain myself clearer!