Maximal Counterexample to Compactness The question is:
Let $X$ be not compact. Prove that there exists an open cover $A$ such that $A$ does not have a finite subcover and any open cover $A \subsetneq B$ has a finite subcover. 
I am stuck on how to start solving this problem. Any suggestions?
 A: Hint: Use Zorn's lemma.
More details are hidden below.

 Consider the poset of open covers of $X$ which have no finite subcover, ordered by inclusion.  All you need to verify is that the union of a chain of such covers is another such cover.

A: Let $\mathcal{U}$ be an open cover of $X$ without a finite subcover. It exists because $X$ is not compact.
As an alternative to Zorn (which is much easier), enumerate all non-empty open sets of $X$ that are not already in $\mathcal{U}$ as $\{O_\alpha: \alpha < \kappa\}$ for some cardinal $\kappa$ (using AC). Define by recursion:
$\mathcal{U}_0=\mathcal{U}$ and having defined $\mathcal{U}_\beta$ for all $\beta < \alpha (< \kappa)$ we define $\mathcal{U}_\alpha = \bigcup_{\beta < \alpha} \mathcal{U}_\beta$ when $\alpha$ is a limit and if not, $\alpha$ is a successor and $\alpha=\beta_0+1$ or $\beta_0 < \alpha$: $\mathcal{U}_\alpha = \mathcal{U}_\beta \cup \{O_{\beta_0}\}$ when the latter cover has no finite subcover and $\mathcal{U}_\alpha = \mathcal{U}_\beta$ otherwise. 
One checks by induction that all $\mathcal{U}_\alpha, \alpha < \kappa$ have no finite subcovers and $\mathcal{A}=\bigcup_{\alpha < \kappa} \mathcal{U}_\alpha$ is your required maximal cover without finite subcover.
But, as said, it's much easier to verify that the set of all covers extending $\mathcal{U}$ that have no finite subcover is inductive (as Bourbaki calls it) because a union of an increasing family of such covers is again of that type, a fact also used in the above induction.
A: By definition, a space S is compact when for all open cover C of S, exists finite subcover Cf subset C with S subset $\cup$ Cf.  
Assume S is not compact.
Thus exists open cover C of S with no finite subcover.  
F = { C open cover S : C does not have a finite subcover }
is not empty.  Pick C in F.
Let K subset F be a chain with C in K.  
C* = $\cup$ K is an open cover of S.
Assume C* has a finite subcover Cf*.
For all U in Cf*, exists K$_U$ in K with U in K$_U.$ 
Since Cf* is finite and K a chain,
exists U in Cf* with Cf* subset K$_U.$
As that contradicts K$_U$ in F, conclude C* in F.  
Thus by Zorn's lemma, F has a maximal element.
