When we assume the continuum hypothesis, for $X=\{0, 1\}$, the $\omega_1$ product $X^{\omega_1} = X^{\mathfrak{c}}$ is not sequentially compact while $X$ is sequentially compact. So the proposition in the title is relative consistent from ZFC. (the assumpution CH can be weaken to $\mathfrak{s} = \aleph_1$, where $\mathfrak{s}$ is the splitting number.)

Then, is it proved in ZFC that for some sequentially compact space $X$, $X^{\omega_1}$ is not sequentially compact?

I suspect that an cardinal $\kappa$ that has $\aleph_1$ cofinality and that is greater than $\mathfrak{c}$ suffices (with the order topology).

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    $\begingroup$ Did you consider $\omega_1^{\omega_1}$? After all, $\omega_1$ is sequentially compact. $\endgroup$
    – Asaf Karagila
    Jan 14 at 15:24
  • $\begingroup$ this question is strictly related to $\mathfrak s$, see the first answer here. As a consequence of that answer you need to consider spaces with big weight as well $\endgroup$ Jan 14 at 21:47
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    $\begingroup$ @AsafKaragila $\omega_1^{\omega_1}$ is sequentially compact under MA and not CH, and not sequentially compact under CH. So it's undecidable. $\endgroup$ Jan 14 at 22:29

According to the Handbook of Set-theoretic Topology (a book that any serious general topologist should have IMO) in the chapters by Vaughn and by van Douwen the fact is mentioned (and partially proved, plus references tot the original papers can be found there as well) that a product of $<\mathfrak{t}$ many sequentially compact (Hausdorff) spaces is again sequentially compact, where $\mathfrak{t}$ is the cardinal invariant known as the tower number (defined on page 115 in the handbook). It is consistent with ZFC that $\mathfrak{c} = \aleph_2 = \mathfrak{t}$ (e.g. by theorem 5.1 of van Douwen's chapter).

In such a model no such $X$ as desired could exist. So the answer is no.


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