Question about exercise 8.5 in Jech's Set theory The exercise I am asking about is the next one:

For every stationary $S\subset \omega_1$ and every $\alpha < \omega_1$ there is a closed set of ordinals A of length $\alpha$ such that $A\subset S$.

The book gives this hint:
By induction on $\alpha$: $\forall \gamma$ $\exists$ closed $A\subset S$ of lenght $\alpha$ such that $\gamma$ < min$A$. The nontrivial step: If true for a limit $\alpha$, find a closed $A \subset S$ of length $\alpha$ such that sup$A\in S$. Let $A_\xi$, $\xi < \omega_1$, be closed subsets of $S$, of length $\alpha$, such that $\lambda_{\xi}$= sup $\cup_{\nu < \xi} A_{\nu}<$min$A_{\xi}$. There is $\xi$ such that $\lambda_{\xi}\in S$. Let $\xi$=lim$_n\xi_n$. Pick initial segments $B_{\xi_n}\subset A_{\xi_n}$ of length $\alpha_n$+1 where lim$_n\alpha_n=\alpha$. Let $A=\cup_{n=0}^\infty B_{\xi_n}$.
There are some things in the hint I don't understand: how do you see that there is  $\xi$ such that $\lambda_{\xi}\in S$. And where do the  $\xi_n$ and $\alpha_n$ come from? 
Can someone help me understanding the hint? Or does someone have any other way of doing it?
 A: Expanding on my comment a bit, what the hint says is that you should show this by induction on $\alpha$. It is also implied by the hint that the limit stage is easy to do, and the non-trivial part is the successor step after a limit (this is because you want to enforce that the set is closed). In fact the hint suggests that you should show more, namely that given $\alpha$ and $\beta$ arbitrary you can find $A\subset S$, such that $\beta<\mathrm{min}A$ and $A$ is a closed set of length $\alpha$.
Now assume that you have that there exists a set of ordinals $B$ that starts arbitrarily high in $\omega_1$, is closed and of length $\alpha$ (where $\alpha$ is a limit). Pick a $\beta<\omega_1$. If you manage to create a new set $A$, subset of $S$, that starts above $\beta$, is closed and of length $\alpha$, and furthermore you have that $\mathrm{sup}A\in S$, then $A\cup\{\mathrm{sup}A\}$ will be a closed subset of $S$ of length $\alpha+1$ that starts above $\beta$.
Take $A_\xi$ sets of length $\alpha$ such that $\beta<\mathrm{min}A_0$ and $\bigcup_{\nu<\xi}A_\nu<\mathrm{min}A_\xi$. This is something you can do by the induction hypothesis. Define $\lambda_\xi=\bigcup_{\nu<\xi}A_\nu$. Observe that the set $\{\lambda_\xi : \xi\in\omega_1\}$ is a closed unbounded subset of $\omega_1$, because if $\xi$ is a limit ordinal $$\lambda_\xi=\bigcup_{\nu<\xi}A_\nu =\bigcup_{\nu<\xi}\bigcup_{\delta<\nu}A_\delta=\mathrm{lim}_{\nu<\xi}\lambda_\nu. $$ Hence $S$ intersects $\{\lambda_\xi : \xi\in\omega_1\}$ (because $S$ is stationary and stationary sets intersect closed unbounded sets), so let $\xi$ be such that $\lambda_\xi\in S$. 
Now take $\xi_n$ be such that $\mathrm{lim}\xi_n=\xi$ and let $\alpha_n$ be such that $\mathrm{lim}\alpha_n=\alpha$. Pick subsets $B_n$ of $A_{\xi_n}$, of length $\alpha_n+1\setminus\alpha_{n-1}$ (we want this $+1$ here to make sure that the union is closed - if we didn't pick successor length, we wouldn't be able to guarantee that the union of these sets would be closed ; in fact it wouldn't be closed by the choice of $A_{\xi_n}$). Then $A=\bigcup_{n\in\omega}B_n$. Notice that the length of $A$ is $\alpha$ and furthermore by definition $\mathrm{sup}(A)=\lambda_\xi$. Hence $A\cup\{\lambda_\xi\}$ is closed, of length $\alpha+1$, subset of $S$ that starts above $\beta$.
