Is the following proof of: $X = [0,\omega_1]$ does not satisfy $S_1(\Omega,\Gamma)$ correct? Definitions:


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*An open cover $\mathcal U$ of $X$ is called a $\gamma$-cover, if for every $x \in X$, the set $\{ U \in \mathcal U : x \notin U \}$ is finite.

*An open cover $\mathcal U$ of $X$ is called an $\omega$-cover, if, for every finite $K \subset X$, there exists $U \in \mathcal U$, such that: $K \subset U$.

*$X$ satisfies $S_1(\Omega,\Gamma)$, if, for each sequence of $\omega$-covers, there exists a sequence $\{ U_n \in \mathcal U_n \}$, such that, $\bigcup U_n$ is a $\gamma$-cover.


I am trying to prove that $X = [0,\omega_1]$ does not satisfy $S_1(\Omega,\Gamma)$:
I tried the following proof. What do you think? does it seems correct?
I claim that:
Claim: $C_p(X)$ where $X = [0,\omega_1]$ is not Frechet-Urysohn.
Proof:
For each finite subset of $K_\alpha \subset [0,\omega_1]$, let, $f_{K_\alpha}(x)=0$ if $x \in K_\alpha$ and 
$f_{K_\alpha}(x) \in (\frac{1}{2},1)$ if $x \notin {K_\alpha}$. Set $A = \{ f_{K_\alpha}\} : K_\alpha$ is a finite subset of $[0,\omega_1] \}$. Then $\overline 0 \in \overline{A \setminus \overline 0}$. on the other hand, any countable set of functions from $A$, has value $0$ on a countable subset  $B \subset [0,\omega_1]$ and has value in $(\frac{1}{2},1)$ for every $x \in B^C$. So, no countable set of functions $C \subset A$ can have $\overline 0$ as an accumulation point.
Corollary:
This also shows that $C_p(X)$ is not Pytkeev, for $X = [0,\omega_1]$
In page 8 of this article, there is a table which summarizes equivalent conditions. Among them is the condition that $Cp(X)$ is FU iff $X$ satisfies $S_1(\omega,\Gamma)$.
What do you think?
Thank you!
 A: Your proof that $C_p([0,\omega_1])$ is not Fréchet–Urysohn is incorrect.
Note that if $f : [0,\omega_1] \to \mathbb{R}$ is continuous, then the set $$\{ \alpha < \omega_1 : f(\alpha) \neq f(\omega_1) \}$$ is at most countable.  (This is because $f^{-1} [ \{ f(\omega_1) \} ]$ must be a Gδ-set containing $\omega_1$, and every open set containing $\omega_1$ is co-countable.)  This means that if $K \subseteq [ 0 , \omega_1 ]$ is finite and $f_K$ is continuous, then $\omega_1 \notin K$.  But then it is easy to see that the constant zero function $\bar{0}$ does not belong to the closure of $\{ f_K : K \subseteq [0,\omega_1)\text{ is finite} \}$.

It turns out that $[0,\omega_1]$ satisfies $\mathsf{S}_1(\Omega,\Gamma)$.

Suppose that $\langle \mathcal{U}_n \rangle_{n \in \omega}$ is a sequence of $\omega$-covers of $[0,\omega_1]$.
  
  
*
  
*Begin by picking $U_0 \in \mathcal{U}_n$ such that $\omega_1 \in U_0$.  Note that $[0,\omega_1] \setminus U_0$ is countable, so enumerate it as $\{ \alpha^0_j \}_{j \geq 0}$.
  
*Pick $U_1 \in \mathcal{U}_1$ such that $\omega_1 , \alpha^0_0 \in U_1$.  Note that $[0,\omega_1] \setminus U_1$ is countable, so enumerate it as $\{ \alpha^1_j \}_{j \geq 1 }$.
  
*Pick $U_2 \in \mathcal{U}_2$ such that $\omega_1 , \alpha^0_0 , \alpha^0_1 , \alpha^1_1 \in U_2$.  Note that $[0,\omega_1] \setminus U_2$ is countable, so enumerate it as $\{ \alpha^2_j \}_{j \geq 2 }$.
  
*In general, pick $U_{n+1} \in \mathcal{U}_{n+1}$ such that $\{ \omega_1 \} \cup \{ \alpha^i_j : i \leq j \leq n \} \subseteq U_{n+1}$, and then note that $[0,\omega_1] \setminus U_{n+1}$ is countable, so enumerate it as $\{ \alpha^{n+1}_j \}_{j \geq n+1}$.
  
  
  Clearly $\{ U_n : n \in \omega \}$ will cover $[0,\omega_1]$, and $\omega_1 \in \bigcap_n U_n$.  Furthermore, given $\alpha < \omega_1$, if there is an $n$ such that $\alpha \notin U_n$, then $\alpha = \alpha^n_i$ for some $i \geq n$, and then $\alpha \in U_m$ for all $m > i$.

