Let $X$ has countable extent. Does $X^2$ have countable extent? 
Definition 1: A space $X$ has countable extent if every uncountable subset of $X$ has a limit point in $X$.

I'm struggling with this question:

Question 2: Let $X$ has countable extent. Does $X^2$ have countable extent? 

It is obvious to see that the answer to the question is NO. The one counterexample is the Sorgenfrey line $S$. It is Lindelof, and hence it has countable extent. However, $S^2$ has uncountable closed discrete subset. 
However I try to prove it is true:), could somebody point out where the proof is wrong?

My proof: Let $A$ be an uncountable subset of $X^2$, say $A=\{(x_\xi,y_\xi):\xi \in \omega_1\}$. Let $A_1=\{x_\xi:\xi\in \omega_1\}$ and 
   $A_2=\{y_\xi:\xi\in \omega_1\}$. $A_1$ or $A_2$ may be countable, however, one of them must be uncountable. 
  
  
*
  
*If $A_1$ and $A_2$ are both uncountable, then $A_1$ has a limit point, say $x$, $A_2$ has a limit point too, say $y$. Then the point $(x,y)$ is the limit point of $A$ in $X^2$.
  
*If only $A_1$ is uncountable, then then $A_1$ has a limit point, say $x$. For $A_2$, there must exist a point $y$ such that $\{(x_\xi,y): \xi\in A_1\}$ must be uncountable. Therefore, we have $(x,y)$ is the limit point of $A$. 
This complete the proof.

Thanks for any help.
 A: Your problem is in case 1, where you assert that if $x$ is a limit point of $A_1 = \pi_1 [ A ]$ and $y$ is a limit point of $A_2 = \pi_2 [ A ]$, then $\langle x,y \rangle$ is a limit point of $A$.
Suppose that your space is the real line $\mathbb{R}$, and suppose you've been strange and your uncountable subset of $\mathbb{R} \times \mathbb{R}$ is the unit circle $S^1 = \{ \langle x , y \rangle \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 = 1 \}$.  Note that $\pi_1 [S^1] = [-1,1]$, and has $0$ as a limit point, and similarly $0$ is a limit point of $\pi_2 [ S^1 ]$.  However $\langle 0 , 0 \rangle$ is clearly not a limit point of $S^1$.
In more concrete terms, taking $\mathbb{R}_{\text{S}}$ to be the Sorgenfrey line, and $A = \{ \langle x , -x \rangle : x \in \mathbb{R} \}$ the anti-diagonal, then every $u \in \mathbb{R}_{\text{S}}$ is a limit point of $\pi_1 [ A ]$, and similarly every $v \in \mathbb{R}_{\text{S}}$ is a limit point of $\pi_2 [ A ]$.  If $v \neq -u$ then it is easy to show that $\langle u , v \rangle$ is not a limit point of $A$ (since $A$ is a closed subset of $\mathbb{R}_{\text{S}} \times \mathbb{R}_{\text{S}}$).  if $v = -u$, then $U = [ u , u+1 ) \times [ v , v+1 )$ is an open neighbourhood of $\langle u , v \rangle$ and $U \cap A = \{ \langle u , v \rangle \}$, so $\langle u , v \rangle$ is not a limit point of $A$.
(Even in case two your proof seems a bit mixed.  If $A_2$ is countable, then there must be a $y \in X$ such that $Z = \{ \xi : \langle x_\xi, y \rangle \in A \}$ is uncountable.  Then the set $\{ x_\xi : \xi \in Z \}$ has a limit point, $x$, and it follows that $\langle x , y \rangle$ is a limit point of $A$.)
