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Let X be compact. Then X is countably compact.
My thinking is like this: Let X be a topological space. Since compact, then every open cover hava a finite subcover. Hence it is true for countable open cover.
Is this proof true?

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    $\begingroup$ a finite set is always countable. $\endgroup$ – Zest Mar 28 at 20:20
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    $\begingroup$ @Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property. $\endgroup$ – MPW Mar 28 at 20:32
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You are exactly correct.

In particular, countable compactness follows from compactness because every countable open cover is an open cover.

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