Every countably compact, scattered $T_3$-space is sequentially compact A space $X$ is called scattered provided every non-empty subspace $Y$ has an isolated point( with respect to the subspace topology on $Y$).
How to prove that:

Every countably compact, scattered $T_3$-space is sequentially compact?

Thanks for your help.
 A: It seems the following. Let $X$ be a countably compact, scattered regular (that is $T_3$ and $T_1$) space and $S=\{x_n:n\in\omega\}\subset X$ be a sequence. If the set $\overline S\backslash S$  is empty then $S$ is closed in $X$ and we put $Y=S$. In the opposite case the subspace $\overline S\backslash S$ has an isolated point $x_0$. Therefore there exists a neighborhood $U_0$ of the point $x_0$ in the space $\overline S$ such that $U_0\subset\{x_0\}\cup S$. Since the space $\overline S$ is regular, there exists an open  neighborhood $V_0$ of the point $x_0$ in the space $\overline S$ such that $\operatorname{cl}_{\overline S} V_0\subset U_0$. Put $Y=\operatorname{cl}_{\overline S} V_0$. In both cases $Y$ is a countable countably compact space, containing  infinitely many members of the sequence $S$ (some of them may coincide as points of $X$). Because $Y$ is a regular countably compact space with $G_\delta$-diagonal, by Chaber   Theorem is a metrizable compact (see, for instance, “Handbook of Set-Theoretic Topology 1984”, p.422 of the file or p.433 of the book). Because $Y$
contains  infinitely many members of the sequence $S$, we can construct a convergent subsequence of $S$ by them.
