Minimal uncountable well-ordered set $S_{\Omega}$ and the sequence lemma Let $S_{\Omega}$ denote an uncountable well-ordered set every section of which is countable.
In Munkres' book Topology, it says on pg 181 that:

$S_{\Omega}$ satisfies the sequence lemma: if $A$ is a subset of $S_{\Omega}$ and $y\in\overline A$, then there exists a sequence $\{a_n\}$ in $A$ such that $a_n \to y$

To prove this, I tried to construct a sequence $\{a_n\}$ by using the fact that the section $\{z\in S_\Omega \mid z < y\}$ is countable, but I am getting no where.
Munkres writes that the claim above can be "readily checked", yet I am stuck. How can I prove this?

Suppose $y\in\overline{A}$. Then first convince yourself that one of the following must hold:


*

*Either $y\in A$ (boring), or

*$y\in \overline{A_y}$, where $A_y$ is the set of elements of $A$ which are $<y$ (if we are thinking of elements of $S_\Omega$ as ordinals, we could write this as "$A\cap y$").
If case one holds, we're done - take $a_n=y$. If case two holds, we need to think a little; we want a sequence of points which "climb all the way up $A_y$." To do this, we're going to use the fact that $A_y$ is countable. List the elements of $A_y$ as $A_y=\{a_0, a_1, a_2, . . .\}=\{a_i: i\in\mathbb{N}\}$; note that this is probably listing them out of order. That's fine. 
Now first, show the following:


*

*For each $n$, we can find some $k$ such that $a_k\ge a_m$ for all $m<n$.


Finally: can we use this to build a sequence which climbs all the way up $A_y$? (HINT: can we ensure that the $n$th term of the sequence be bigger than each of $a_0, a_1, a_2, . . . , a_n$?)
