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3 added 60 characters in body

Le $X$ be any Hausdorff, sequentially compact, not compact space (e.g. $\omega_1$ with the order topology). Then $X^\mathbb{N}$ is Hausdorff, sequentially (hence countably) compact, and not locally compact, because any set with non-empty interior is mapped surjectively onto $X$ by some projection, thus is not compact as $X$ itself is not.

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Le $X$ be any Hausdorff, sequentially compact, not compact space (e.g. $\omega_1$ with the order topology). Then $X^\mathbb{N}$ is Hausdorff, sequentially compact, and not locally compact, because any set with non-empty interior is mapped surjectively on onto $X$ by some projection.

1

Le $X$ be any Hausdorff, sequentially compact, not compact space (e.g. $\omega_1$ with the order topology). Then $X^\mathbb{N}$ is Hausdorff, sequentially compact, and not locally compact, because any set with non-empty interior is mapped surjectively on $X$ by some projection.