Example 3, Sec. 28 in Munkres' TOPOLOGY, 2nd ed: How does $S_\Omega$ satisfy the sequence lemma? Here's the sequence lemma:

Let $X$ be a topological space, let $x \in X$, and let $A \subset X$. If there is a sequence of points of $A$ converging to $x$, then $x \in \overline{A}$. Conversely, if $X$ is metrizable and if $x \in \overline{A}$, then there is a sequence of points of $A$ converging to $x$.

Then how to show that the minimal uncountable well-ordered set $S_\Omega$ satisfies the sequence lamma?
Here's how $S_\Omega$ evolves:

Let $B$ be any uncountable well-ordered set, and let $C \colon= \{1, 2 \} \times B$ in the dictionary order. Then the set $C$ is well-ordered. Let $C_0$ be the subset of $C$ consisting of those elements $a$ of $C$ by which the section $S_a$ of $C$ is uncountable. For any $a \in C$, we define $S_a$ as follows: $$S_a \colon= \{ \ x \in C \ \colon \ x < a \ \}.$$ Then $C_0$ is non-empty. In fact, any element of the form $2 \times b$, where $b \in B$, belongs to $C_0$. So $C_0$ has a smallest element $\Omega$. Thus, $S_\Omega$ is uncountable, but for any element $a$ of $S_\Omega$, the section $S_a$ is countable. The topologies on $S_\Omega$ or $\overline{S_\Omega}$ are the respective subspace topologies these inherit as subsets of the set $C$, or their respective order topologies.

Also, $\Omega$ is a limit point of $S_\Omega$, but there is no sequence in $S_\Omega$ that converges to $\Omega$; in fact, any countable subset of $S_\Omega$ has an upper bound in $S_\Omega$. Thus, $\overline{S_\Omega} = S_\Omega \cup \{ \Omega \}$ does not satisfy the sequence lemma.
PS:
My Attempt:

Let $A$ be a non-empty subset of $S_\Omega$, and let $p \in S_\Omega$ be an adherent point of set $A$. We need to show that there exists a sequence $\left(p_n\right)_{n \in \mathbb{N}}$  in $S_\Omega$ that converges to point $p$.
If $p \in A$, then we can take $p_n \colon= p$ for all $n \in \mathbb{N}$, and we are done. So let us suppose that $p \not\in A$. Then $p$ is of course a limit point of $A$.

What next? How then can we construct a sequence $\left( p_n \right)_{n \in \mathbb{N} }$ that converges to $p$?
 A: Let $a\in S_\Omega$. Then $S_a$ is countable, so $\{a\}\cup S_a$ is countable, and $S_\Omega\setminus(\{a\}\cup S_a)$ is uncountable and therefore non-empty. $S_\Omega$ is well-ordered, so $S_\Omega\setminus(\{a\}\cup S_a)$ has a least element; call it $a^+$. Suppose that $U$ is an open nbhd of $a$ in $S_\Omega$.
Case 1. Suppose first that this $a$ is NOT the smallest element of $S_\Omega$.
Then by definition there are $b,c\in S_\Omega$ such that $a\in(b,c)\subseteq U$, where $(b,c)=\{x\in S_\Omega:b<x<c\}$. Clearly $a<c$, so $a^+\le c$, and
$$a\in(b,a^+)\subseteq(b,c)\subseteq U\;.$$
Thus, $\mathscr{B}_a=\{(b,a^+):b\in S_a\}$ is a local base at $a$, and since $S_a$ is countable, so is $\mathscr{B}_a$.
Case 2. Next, suppose that $a$ is the smallest element of $S_\Omega$. Then there exists an element $c \in S_\Omega$ such that $a \in [a, c) \subset U$. Thus we again have $a^+ \leq c$, and so $a \in \left[a, a^+ \right) \subset [a, c) \subset U$.
Therefore we have the following as a local base at $a$:
$$ \mathscr{B}_a = \left\{ \ \left[ a, a^+ \right) \ \right\}. $$
This shows that $S_\Omega$ is first countable. The desired result now follows from the following lemma.

Lemma. If $X$ is a first countable space, $x\in X$, $A\subseteq X$, and $x\in\operatorname{cl}A$, then there is a sequence in $A$ converging to $x$.
Sketch of Proof. Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a countable local base at $x$. For each $n\in\Bbb N$ let $U_n=\bigcap_{k\le n}B_k$; $U_n$ is an open nbhd of $x$, so there is a point $x_n\in U_n\cap A$. Now check that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$. $\dashv$

