if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$ is compact or......... 
Theorem: if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$  is compact or  there is a sequence  in $A$ converging to point  of $ X- A $.
proof: Suppose $ A‎\subset‎  X$ is not compact. Let  $D$ be an infinite subset  of $S$ which  is closed  in $A$ . since $X$  is compact , $ D ^{d}‎ \neq‎ ‎\emptyset $ and $ D ^{d} \subseteq X - A $. we enumerate $ \text{cl} (D) - D $ as $ \{   x_{n}  :  n \in \omega  \}$ and we will show  that for some $n$, $x_{n}$  is the  limit  of  a sequence  in $D$ , showing  that $A$  is not  sequentially closed.
If each neighborhood $U$  of $x_{0} =  z_{0} $ is such that $ D - U $ is finite., then any  enumeration  of $D$  to $ z_{0} $ . If not,  then pick an open  set $ U_{0}$  such that  $ D - U_{0} $ is  infinite and  $ z_{0} \in U_{0} $ : note that  since $X$  is compact  and $D$ is discrete,$ ( \text{cl}( D) - D ) - U_{0} \neq‎ \emptyset$‎ . Now let  ‎$ z‎_{1} =‎ ‎x‎_{m‎_{1}‎} $ , where $ m_{1} = \inf \{ n \in \omega : x_{n} \not\in U_{0} \} $. If  each neighborhood $ U$  of $z_{1} $ is  such that $(  D - U_{0} ) - U $ is finite, then  any enumeration of $ D - U_{0} $ will converge to $z_{1}$. Having chosen  point $ z_{0},z_{1}....z_{k-1}  $ and open  sets containing them $U_{0}, ...,.U_{k-1} $ in such that  $ D - ‎\bigcup‎ \{ U_{j} ; 0 ‎\leq‎ j ‎\leq‎ k-1 \} $ is infinite, it is  clear as  before  that $ ( \text{cl}( D) - D ) - ‎\bigcup‎ \{ U_{j} ; 0 ‎\leq‎ j ‎\leq‎ k-1 \}  $ is not  empty and We  let $z‎_{k} =‎ ‎x‎_{m‎_{k}‎}  $ where $ m_{k} = \inf \{ n \in \omega : x_{n} \not\in  ‎\bigcup‎ \{ U_{j} ; 0 ‎\leq‎ j ‎\leq‎ k-1 \} \}$. As before , either every neighborhood $ U$  of $z_{k} $ is such that $ ( D - U ) - \bigcup‎ \{ U_{j} ; 0 ‎\leq‎ j ‎\leq‎ k-1 \} \}$ is finite( in which  case  we obtain a sequence convergent to $ Z_{k} $ ) or there  is some $ U = U_{k} $  for which  this set is infinite.
However , since $D$ is locally compact , $ \text{cl}(D) - D $ is compact and hence  for some $ n \in \omega  $,$ (\text{cl}( D) - D ) - ‎\bigcup‎ \{ U_{j} ; 0 ‎\leq‎ j ‎\leq‎ n \} =   ‎\emptyset$. but ‎$ (\text{cl}( D) - D ) - ‎\bigcup‎ \{ U_{j} ; 0 ‎\leq‎ j ‎\leq‎ n -1 \} ‎\neq‎   ‎\emptyset$. It is then the case that any enumeration of $ D - ‎\bigcup‎ \{ U_{j} ; 0 ‎\leq‎ j ‎\leq‎ n -1 \} $ will converge to $ z_{n} $.
But:
(a) : why in the first paragraph, We say that $D$ is discrete , and closed in $A$? In this case, is $x_{0} $ only accumulation point of $D$?
(b) : why in the second part, we say " If each neighborhood $U$  of $x_{0} =  z_{0} $ is such that $ D - U $ is finite., then any  enumeration  of $D$  to $ z_{0} $ "? and  why $ ( \text{cl}( D) - D ) - U_{0} \neq‎ \emptyset$ ?
(c): why  in the last part,  $D$ is locally compact and   $ \text{cl}(D) - D $ is compact ?

 A: The first paragraph doesn't say that $D$ is discrete, but it should. The point is that $A$ is not compact, so it not countable compact and therefore must contain an infinite closed discrete subset $D$. However, $X$ is compact, so $D$ must have a limit point in $X$. This means that $D^d=(\operatorname{cl}_XD)\setminus D$ cannot be empty. However, $D$ has no limit points in $A$, so $D^d\subseteq X\setminus A$.
$D^d$ might be countably infinite, or it might be finite. The copied proof assumes the former case, but I won't: there is an $\alpha\le\omega$ such that we can enumerate $D^d=\{x_k:k<\alpha\}$. It's possible that $\alpha=1$ and that $x_0$ is the only limit point of $D$, but this certainly need not be the case. At any rate, let $z_0=x_0$. There are two cases:


*

*For every open nbhd $U$ of $z_0$, $D\setminus U$ is finite. As we saw in another of your questions, this means that no matter how we enumerate $D=\{y_n:n\in\omega\}$, the sequence $\langle y_n:n\in\omega\rangle$ converges to $z_0$. And in that case we're done: $\langle y_n:n\in\omega\rangle$ is a sequence in $A$ converging to a point of $X\setminus A$.

*There is an open nbhd $U_0$ of $z_0$ such that $D\setminus U_0$ is infinite. Since $X$ is compact, the infinite set $D\setminus U_0$ has a limit point in $X$, and since $D\setminus U_0$ is closed and discrete in $A$, this limit point cannot be in $A$: it must be in $D^d$. It's certainly not in $U_0$, so $D^d\setminus U_0\ne\varnothing$. Thus, there is a smallest $m_1<\alpha$ such that $x_{m_1}\in D^d\setminus U_0$, and we let $z_1=x_{m_1}$ and $D_1=D\setminus U_0$.
In the second case we repeat the argument, using $z_1$ and $D_1$ instead of $z_0$ and $D$. There are again two cases:


*

*For every open nbhd $U$ of $z_1$, $D_1\setminus U$ is finite. Then any enumeration of $D_1$ converges to $z_1$, and we're done: any enumeration of $D_1$ is a sequence in $A$ converging to a point of $X\setminus A$.

*There is an open nbhd $U_1$ of $z_1$ such that $D_1\setminus U_1$ is infinite. As before, $D_1\setminus U_1$ has a limit point in $D_1^d\setminus U_1$, so there is a smallest $m_2<\alpha$ such that $x_{m_2}\in D_1^d\setminus U_1$, and we let $z_2=x_{m_2}$ and $D_2=D_1\setminus U_1$.
In this way we continue the recursive construction of the points $z_k$. The construction terminates only if at some point we're in case (1), so the proof will be complete if we show that it must terminate in a finite number of steps. This is what is done in the last paragraph of the copied proof.
$D$ is a discrete subspace of $X$, and a discrete space is trivially locally compact: for each $x\in D$, $\{x\}$ is a compact open nbhd of $x$ in $D$. A locally compact space is open in any compactification, and $\operatorname{cl}_XD$ is certainly a compactification of $D$, so $D$ is open in $\operatorname{cl}_XD$. It follows that $D^d=(\operatorname{cl}_XD)\setminus D$ is closed in $\operatorname{cl}_XD$ and therefore compact.
