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Take $X$ a Tychonoff product $[0,1]^{\omega_1}$ and as $Y$ the $\Sigma$-product $$ \{ x ∈[0,1]^{\omega_1}:|\{\alpha<\omega_{1} :x(\alpha)\ne 0 \}|\le\omega \}\;.$$ The space $X$ is compact by Tychonoff Theorem and $Y$ is a countably compact and not closed subspace of $X$.

why $Y$ is a countably compact and not closed subspace of $X$?

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The subspace $Y$ is not closed; in fact, it is dense in $X$. Already the subspace $Z$ consisting of all those elements which are zero except at finitely many points is dense in $X$. If $U$ is a basic open subspace which is nonempty then it can only put restrictions on finitely many coordinates; say $U$ is of the form $\bigcap {\pi^{-1}_{\alpha_i}}(U_{\alpha_i})$ then every $x$ such that $x(\alpha_i)$ is in $U_{\alpha_i}$ and zero otherwise lies in both $Y$ and $U$.

(In fact, $Y$ is not compact and therefore cannot be closed in $X$ - can you see why?)

To see that the space $Y$ is countably compact it is sufficient to prove that every countable set has an accumulation point. So let $A$ be a countable subset of $Y$. The set $A$ has "countable support" - that is, there is a countable ordinal $\beta$ such that all elements of $A$ vanish beyond the $\beta$-th coordinate. Now, $A$ is naturally also a subspace of $[0,1]^{\beta}$, which is compact. Find an accumulation point $y$ of $A$ in this space and extend it to zero after the $\beta$-th coordinate; this gives an accumulation point of $A$ in $Y$.

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  • $\begingroup$ @Habib: You can find a more extensive discussion here in Dan Ma's Topology Blog. $\endgroup$ – Brian M. Scott Aug 12 '13 at 13:46

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