topological sequential space $(X,\tau)$ Suppose in topological space $(X,\tau)$ every countably compact is closed.Let  $(X,\tau)$ be sequential space.
(1): If every  infinite subset $A \subseteq X$ is closed, will $A$  be discreet in $X$?
(2)If there is $x \in \overline{C} - C$ and $\{ x_n :n \in \omega\}$ in infinite and non-closed $ C \subseteq A$ s.t $\{ x_n :n \in \omega\}$ converge to $x$,  won`t $x_n$ be an accumulation point for $\{ x_n :n \in \omega\}$?why?
 A: *

*Yes. This is true in general, not just for sequential strongly $KC$ spaces: if $A$ is an infinite subset of a space $X$, and every infinite subset of $A$ is closed in $X$, then $A$ is a closed, discrete subset of $X$. $A$ is an infinite subset of itself, so $A$ is closed in $X$. For each $x\in A$ the set $A\setminus\{x\}$ is an infinite subset of $A$, so it’s closed in $X$, and $\{x\}=A\setminus(A\setminus\{x\})$ is therefore open in $A$. Thus, for each $x\in A$ there is an open $U_x$ in $X$ such that $U_x\cap A=\{x\}$, and $A$ is discrete as well as closed in $X$.

*Now suppose that $X$ is a sequential $KC$ space, and that $C$ is an infinite subset of $X$ that isn’t closed in $X$. Since $X$ is sequential, $C$ is not sequentially closed, and there are therefore an $x\in(\operatorname{cl}C)\setminus C$ and a sequence $\langle x_n:n\in\omega\rangle$ in $C$ converging to $x$. Let $S=\{x_n:n\in\omega\}$, and let $K=\{x_n:n\in\omega\}\cup\{x\}$; then $K$ is compact, and since $X$ is $KC$, $K$ is closed in $X$. Let $y\in X$. If $y\notin K$, then $X\setminus K$ is an open nbhd of $y$ disjoint from $S$, so $y$ is not an accumulation point of $S$. Suppose now that $y=x_m$ for some $m\in\omega$; the subsequence $\langle x_n:n>m\rangle$ still converges to $x$, so $\{x\}\cup\{x_n:n>m\}$ is compact and therefore closed in $X$. Let $U=X\setminus\big(\{x\}\cup\{x_n:n>m\}\big)$; $U$ is open in $X$. $X$ is $T_1$, so $\{x_n:n<m\}$ is closed in $X$, and therefore $U\setminus\{x_n:n<m\}$ is an open nbhd of $x_m$. Finally, $U\cap S=\{x_m\}$, so $x_m$ is not an accumulation point of $S$. Thus, $x$ is the unique accumulation point of $S$.
These two points together show that every sequential $KC$ space has the $FDS$ property; the space need not be strongly $KC$. Let $A$ be any infinite subset of such a space. If $A$ has an infinite subset $C$ that is not closed in $X$, part (2) shows that $C$ (and therefore $A$) has an infinite subset $S$ with only one accumulation point. If every infinite subset of $C$ is closed in $X$, part (1) shows that $C$ is closed and discrete and therefore has no accumulation points.
