the minimal uncountable well-ordered set $S_\Omega$ and the sequence lemma (example 3, sec 28 in Munkres topology) 
First
How do prove that $S_\Omega$ satisfies the sequence lemma? Munkres says 'you can readily check', but it is not easy for me.
Second
How does the fact that there is no sequence of $S_\Omega$ converging to $\Omega$ imply that $\overline{S_\Omega}$ does not satify the sequence lemma? I understand that there is no sequence of $S_\Omega$ converging to $\Omega$. But $\overline{S_\Omega}$ contains $\Omega$, so if $x_n=\Omega$ for each positive integer $n$, is not $(x_n)$ a sequence of $\overline{S_\Omega}$ converging to $\Omega$? If I'm right, I think we should try a different counterexample (or proof)
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 $x$.

 A: For the second question, the constant sequence for $\Omega$ is not a sequence of points in $S_\Omega$, because $\Omega$ does not belong to that set.
If you replace $A$ with $S_\Omega$, $x$ with $\Omega$, and $X$ with $S_\Omega \cup \{\Omega\}$ in the statement of the sequence lemma, you will see that it fails the converse. If $S_\Omega \cup \{\Omega\}$ was metrizable, we could choose a sequence $(x_n)_{n\in\mathbf{N}}$ (whose points are in $S_\Omega$) that converges to $\Omega$. It has a strictly increasing subsequence $(y_n)_{n\in\mathbf{N}}$, and $\Omega$ is the least upper bound of this subsequence. The set $\bigcup_{n\in\mathbf{N}} S_{y_n}$ is a countable union of countable sets contained in $S_\Omega$, so it cannot be equal to the uncountable $S_\Omega$. This implies that there is a $B$ in $S_\Omega$ which is not in this union, and this $B$ will be an upper bound for our subsequence. It follows that $B < \Omega$, but this contradicts the fact that $\Omega$ is the least upper bound for this subsequence. Hence, the space $S_\Omega \cup \{\Omega\}$ is not metrizable.
Mutatis mutandis, the preceding argument is exactly the reason why the space $S_\Omega$ does satisfy the sequence lemma, and that should settle the first question. If $x$ is a limit point of a subset $A$, then it cannot be the immediate successor of something in $S_\Omega$, and $S_x$ must be (countably) infinite. Can you find a sequence in $A$ that converges to $x$? (Hint: Let $y$ be the immediate successor of $x$. Use a bijection from $\mathbf{N}$ to $S_x$ to construct a strictly increasing sequence $(\alpha_n)_{n\in\mathbf{N}}$ whose least upper bound is $x$, and choose $x_n$ in the intersection of $(\alpha_n, y)$ with $A\setminus \{x\}$.)
