$\{A_\alpha:\alpha<\aleph_1\}$ and $A_\alpha\subseteq\alpha$. Prove existence of stationary $T$ such that $\bigcup_{\alpha\in T}A_\alpha$ is finite. We have family of finite sets $\{A_\alpha:\alpha<\aleph_1\}$ such that $A_\alpha\subseteq\alpha$ for $\alpha<\aleph_1$. Prove that there exists set $T\in Stat_{\aleph_2}$ such that $\bigcup_{\alpha\in T}A_\alpha$ is finite.
I should use Fodor Lemma here somehow. I tried to take set $B$ of ordinals with $\aleph_0$ cofinality and define regressive function $f(\alpha)=\sup A_\alpha$ (it is regressive) for $\alpha\in B$, but I do not see that stationary set resulting from such function is desired one.
 A: I guess there are several ways to skin this cat but this one came to mind first: Write, for each $\alpha < \omega_1$, $$A_{\alpha} = \{x^{\alpha}_1 < \ldots < x^\alpha_{n_{\alpha}} \}$$
Let
$$
f \colon \omega_1 \to \omega_1, \alpha \mapsto \langle n_{\alpha} , \langle x^{\alpha}_1, \ldots , x^\alpha_{n_{\alpha}} \rangle \rangle,
$$
where $\langle . \rangle$ is the (nested) Gödel pairing function.
Let $S \subseteq \omega_1$ be the stationary set of infinite ordinals closed under the Gödel pairing function. $f \restriction S$ is regressive. Hence, by Fodor, there is some stationary $T \subseteq S$ and some $\beta$ such that $f \restriction T = T \times \{ \beta \}$. $\beta$ now codes a unique, finite set $A \subseteq \omega_1$ via the procedure above $(\dagger)$ and we obtain that $A_{\alpha} = A$ for all $\alpha \in T$. In particular $\bigcup_{\alpha \in T} A_{\alpha} = A$ is finite.

$(\dagger)$ There are unique $n < \omega$ and $x_1 < \ldots < x_n < \omega_1$ such that $\langle n, \langle x_1, \ldots, x_n \rangle \rangle = \beta$ and then $A = \{x_1, \ldots, x_n \}$ is the set coded by $\beta$. We need to include $n$ in this coding because we need to know how often we nested the Gödel pairing function in the second coordinate. Otherwise different sequences of necessarily different lengths will get mapped to the same ordinal -- ruining our coding.
A: Here's a "coding-free" alternative to Stefan's answer.
First, a quick fact:

Claim. Suppose I partition $\omega_1$ into countably many pieces. Then (at least) one piece is stationary.

Proof: if $P_i$ is nonstationary for each $i\in\omega$, let $C_i$ be a club disjoint from $P_i$; then $\bigcap C_i$ is a club disjoint from all $P_i$s, so we can't have $\bigcup_{i\in\omega} P_i=\omega_1$. $\quad\Box$
From the claim, we now argue as follows:
Given $\{A_\alpha:\alpha\in\omega_1\}$ as above, let $P_i=\{\alpha:\vert A_\alpha\vert=i\}$. By the claim above, for some $i$ we have that $P_i$ is stationary. So fix such an $i$.
Now for $n<i$ let $g_n$ be the function, with domain $P_i$, given by: $g_n(\alpha)$ is the $n$th element of $A_\alpha$ (counting from least to greatest). Each $g_n$ is regressive, and so we can iterate Fodor's lemma to get a stationary $S\subseteq P_i$ on which each $g_n$ is constant. But this means exactly that $A_\alpha=A_\beta$ for each $\alpha,\beta\in S$.
