Uncountable subset of first uncountable ordinal set Given $g: \omega_1\rightarrow \omega_1$ is a function such that if $x\neq 0$, then $g(x)<x$ ($g$ is not necessarily continuous). Prove that there exists $t\in \omega_1$ such that $\ f^{-1}(t)$ is uncountable
My attempt By using Pressing-down lemma, we know that there exists a stationary subset $\ K\subset \omega_1$ such that $f(K)$ is constant. Since $K$ = subset that has nonempty intersection with EVERY closed and unbounded sets in $\omega_1$, and there are uncountable such closed and unbounded sets, there must be uncountable intersections with $K$. The union of such uncountable intersections is uncountable , so $K$ must be uncountable. 
Can anyone help review my proof above to see if it has any error?
 A: Essentially you have to prove that a stationary set is unbounded, and therefore uncountable (in the case of $\omega_1$, at least).
The argument is a bit murky. You should explicitly point out that the tail segments are closed and unbounded sets; so a stationary set meets every tail segment. It might be slightly easier to prove that a bounded set cannot be stationary instead.

Here is an example for a fully written proof of this statement (as asked by the OP in the comments).
By the pressing-down lemma, there exists a stationary subset $K\subseteq\omega_1$ such that $f(K)$ is constant. It suffices to show that $K$ is uncountable. For this we will show that if $S\subseteq\omega_1$ is countable, then it is not a stationary set.
Suppose that $S$ is countable, then $\bigcup S$ is the countable union of countable ordinals, and therefore1 $\bigcup S$ is a countable ordinal itself, denote it by $\alpha$. Consider now the set $C=\{\gamma<\omega_1\mid\alpha<\gamma\}$, then $C\cap S=\varnothing$, moreover $C$ is a tail segment and so it is closed and unbounded.2 Therefore $S$ is not stationary.
It follows, if so, that since $K$ is stationary it must be uncountable. $\square$



*

*Here we use the axiom of choice. It might be the case that $\omega_1$ is the countable union of countable sets. But since we already use the pressing-down lemma, we made an appeal to a sufficient amount of choice to begin with.

*I'm assuming that you proved that tail segments are clubs. If you have not, it might be a good exercise.
