The space $\omega_1$ with its order topology J. Van Mill in here http://www.sciencedirect.com/science/article/pii/S0166864107000193 shows that the space $\omega_1$ with its order topology satisfies the following condition: for any neighbourhood assignment $\{O_{\alpha}: \alpha < \omega_1\}$ of the space $\omega_1$, there exists a discrete subset $A$ of $\omega_1$ such that $\bigcup_{\alpha \in A} O_{\alpha} = \omega_1$.
The proof as follows: Take any neighbourhood assignment $\{O_{\alpha}: \alpha < \omega_1\}$ of the space $\omega_1$. For any non-isolated point $\alpha \in \omega_1$, there is $f(\alpha) < \alpha$ such that $(f(\alpha), \alpha] \subset O_{\alpha}$. By pressing down lemma, we can find an uncountable $A\subset \omega_1$ and $\beta <\omega_1$ such that $f(\alpha) = \beta$ for any $\alpha \in A$.
Note: A neighbourhood assignment in a space $X$ is a family $\{O_x: x \in X\}$ such that $x \in O_x \in \tau(X)$ for any $x \in X$.
I do not understand this proof. Why is  $\bigcup_{\alpha \in A} O_{\alpha} = \omega_1$? and why is $A$ discrete? Please help me to understand this proof.
 A: The proof goes on: as $\omega_1$ is scattered, there is a discrete uncountable $B \subseteq A\setminus (\beta+1)$. it is clear that $\cup_{\alpha \in B} O_\alpha \supseteq (\omega_1) \setminus (\beta+1)$ and since $\beta+1$ is compact, we pick a finite subcover of its asignment cover indexed by some finite $F \subseteq \beta+1$. The claim then is that $F \cup B$ is as required (discrete) etc. 
better link: from author's homepage [PDF]
A: The subspace $A$ may fail to be discrete.  Let $$B=\{\min (A \backslash (x+1) : x\in A\}.$$ Then $B$ is an uncountable discrete subspace of $\omega_1,$ and  $\min B >a.$ For $\min B< y<\omega_1$ there exists $x\in B$ \ $(y+1)$ , so $  O_x\supset (a,x]\supset \{y\}.$ Hence $$(\min  B,\omega_1) \subset \cup_{x\in B} O_x.$$ Since the subspace $S=(\min B)+1 $ is compact, there is a finite $C\subset S$ with $$\cup_{c\in C}O_c\supset S.$$ Since $C$ is finite (and hence discrete) and is disjoint from the closure (in $\omega_1$) of $B,$ the space $B\cup C$ is discrete. And $\cup_{x\in B\cup C}O_x=\omega_1.$ 
