Compactness implies countably compactness

Question

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?

Answer 1

You are exactly correct.

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