Open cover of non-compact spaces Let $X$ be a non-compact space. (A space is compact if any open cover has a finite subcover.) I want to show that there is an ordinal $\alpha$ and an open cover $(U_\xi)_{\xi < \alpha}$ such that if we let
$$A_\xi = X \setminus \bigcup_{\eta < \xi} U_\eta $$
for any $\xi < \alpha$, then $\mathrm{Card}(A_\xi)=\alpha$.
I really can't see where to begin. Could someone give me a clue? 
Thank you.
 A: (not a general proof, but a special case:) if you have known that $X$ is not sequentially compact space, you may take a sequence $x_n, n\in\mathbb N,$ without convergent subsequence and let $U_n:=X\setminus \{x_n, x_{n+1}, x_{n+2}\dots\}$.
Noah Schweber showed how to generalize this idea, but somehow he deleted his answer. Here is some about his proof:
For some initial ordinal $\alpha$ take infinite cover $(V_\xi)_{\xi<\alpha}$ such that 
it hasn't subcover of less cardinality, and 
$V_{\xi_0}\nsubseteq\bigcup_{\xi<\xi_0}V_\xi$ for each $\xi_0$. 
For existance such cover use transfinite induction.
Denote by $\beta$ such ordinal that $|X\setminus\bigcup_{\xi<\beta}V_\xi|>\alpha$ but  $|X\setminus\bigcup_{\xi\le\beta}V_\xi|\le\alpha$. $ $ Let $U_\eta:=\bigcup_{\xi<\beta+\eta}V_\xi$.
If for some ordinal $\gamma$ we have $|X\setminus\bigcup_{\xi<\gamma}V_\xi|<\alpha$, than $\alpha$ is not initial ordinal (because $\gamma<\alpha$ but $|\gamma|=|\alpha|$). So we have for each $\xi$ that $|X\setminus\bigcup_{\eta<\xi}U_\eta|=\alpha$.
