TITLE: A problem on sequentially compact and countably compact
QUESTION [6 upvotes]: Recently I came across a problem as follows
"I know that sequentially compact implies countably compact. But can anybody tell me please that the converse is true or false."

REPLY [3 votes]: The converse is false. In fact, even compactness doesn’t imply sequential compactness: see Example $3$ in this PDF. (The space in question is the product of $2^\omega=\mathfrak c$ copies of the discrete $2$-point space.)

REPLY [2 votes]: Even compactness does not imply sequential compactness.  Consider the Stone–Čech compactification of $\mathbb N$, $\beta \mathbb N$.  This space is (by definition) compact, but has no non-trivial (i.e., not eventually constant) convergent sequences, and so any one-to-one sequence does not have a convergent subsequence.