Is there a sequential space that is compact but not sequentially compact? A space $X$ is


*

*sequential if every sequentially closed set is closed,

*compact, if every open cover has a finite subcover,

*sequentially compact, if every sequence has a convergent subsequence.


Any first-countable space is sequential. Any first-countable compact space is sequentially compact.
The typical example $I^I$ with $I = [0,1]$ for a compact not sequentially compact space is not sequential. For such a counterexample it is necessary that $X$ is not first-countable.
The database version of Steen-Seebach $\pi$-base does not provide such an example.

The equivalence of countable compactness and sequential compactness for sequential spaces is true only when the space is $T_1$.
I have seen the proof, which has a bug and the end part, because   $x$ being cluster point of $(x_n)_{n\ge N}$ is not always true, and
if the space is $T_1$, you can say that.
The condition of $T_1$ is necessary
