If X is a countable KC-space, then every infinite $D\subset X$ contains an infinite subset with only a finite number of accumulation points (in X) 
a topological space is called KC- space if every compact set is closed.
a topological space is called US, if every convergent sequence has unique limit.
  generally, KC- space imply US - space.
I have read bellow theorem from " Spaces in which compact subsets are closed and the lattice of T1-topologies on a set " by " Ofelia T. Alas, Richard G. Wilson ", dml.cz link.
But I have several difficulty with this. He said that:
" Lemma 8. If X is a countable KC-space, then every infinite $ D ‎‎‎\subseteq ‎X‎ $‎‎   contains an infinite subset with only a finite number of accumulation points (in $ X $ )."
" Proof: Enumerate $ X $ as $ \{ ‎x‎_{n} : n ‎‎\in ‎\omega ‎‎‎‎‎\} $ and suppose that $ D ‎‎‎\subseteq ‎X‎ $‎‎  is infinite and every infinite subset of $ D $  has infinitely many accumulation points. Let $ ‎n‎_{0} ‎‎‎\in ‎\omega ‎‎‎$‎ be the smallest integer such that $x‎_{n‎_{0}‎}‎ $ is an accumulation point of $ D $  . If each neighborhood V of $x‎_{n‎_{0}‎}‎ $  has the property that $ D ‎\setminus ‎V‎ $‎ is finite, then any enumeration of $ D $   converges to $x‎_{n‎_{0}‎}‎ $‎ and hence $ D $   has only one accumulation point, a contradiction. Thus we may choose an open neighborhood $ V‎_{0}‎  $ of  $ x‎_{n‎_{0}‎}‎ $ such that $ ‎D‎_{1} =‎ D‎ ‎‎\setminus ‎V‎_{0}‎‎ $‎ is infinite. Having chosen points ‎$ ‎x‎_{n‎_{0}‎}‎,‎‎‎ ‎x‎_{n‎_{‎1‎}‎}, ‎\ldots , ‎‎‎‎‎x‎_{n‎_{‎j-1‎}‎} ‎‎$‎ and open sets $ ‎V‎_{0}‎,‎ ‎V‎_{‎1‎‎}, ‎\ldots , ‎V‎_{‎j-1‎‎}‎‎‎$ such that ، ‎‎‎$ x‎_{n‎_{k}‎}‎ ‎\in ‎V‎_{k}‎‎ $‎‎ for each ‏‎$ 1‎ ‎‎\leq k‎ ‎‎\leq ‎j-1‎ $‎‎ and $ ‎D‎_{j} = D ‎‎\setminus \big( ‎\bigcup \{ ‎V‎‎_{k} : ‎1‎\leq k ‎\leq j-1‎‎‎‎‎‎\} \big)‎‎‎‎‎$‎ is infinite, we let ‎$ ‎n‎_{j}‎‎ $ be the least integer such that $ ‎x‎_{n‎_{j}‎}‎‎ $‎  is an accumulation point of $ D‎_{j}‎ $‎ and we choose a neighborhood ‎$ ‎V‎_{j}‎‎ $ of $ ‎x‎_{n‎_{j}‎}‎‎ $‎   such that $ ‎D‎_{j+1} = ‎D‎_{j } ‎\setminus ‎V‎_{j} ‎‎‎‎‎‎$ is infinite. Such a choice is again possible for if every neighborhood V of $ ‎x‎_{n‎_{j}‎}‎‎ $‎    is such that $ ‎D‎_{j}‎‎ ‎\setminus ‎V‎ $ is finite, then any enumeration of $ D‎_{j}‎ $ is a sequence which converges to $ ‎x‎_{n‎_{j}‎}‎‎ $‎    and hence $ D‎_{j}‎ $ has only one accumulation point.
Now for each $ j ‎\in‎ ‎\omega‎ $ we choose $ ‎y‎_{j} ‎\in ‎‎‎ ‎D‎_{j}‎‎ ‎\setminus ‎\{ ‎y‎_{0}‎,‎‎‎ ‎y‎_{‎1‎‎},‎ ‎\ldots,‎ ‎y‎_{‎j-1‎‎}‎ ‎‎ ‎\}‎ $ and we denote the set $ ‎\{ ‎y‎_{n} : n ‎‎\in‎\omega‎‎‎‎‎‎\}‎ $ ‎‎by $ S $. It is clear that $ S $.  is infinite and all but finitely many points of S are contained in $ D‎_{j}‎ $ for each $ j‎ ‎‎\in‎‎ ‎\omega‎ $‎ and so an accumulation point of S is an accumulation point of $ S‎ ‎\cap ‎D‎_{j}‎‎ $‎ for each $ j‎ ‎‎\in‎‎ ‎\omega‎ $‎. Thus $ S $.  can have no accumulation point, since if p were such a point, then for some $ k‎ ‎‎\in‎‎ ‎\omega‎ $,  $ p‎ =‎ ‎x‎_{k}‎‎ $ and from the construction, we would have that ‎‎$ k‎ ‎‎\geq ‎n‎_{j}‎‎ $ for each $ j‎ ‎‎\in‎‎ ‎\omega‎ $, which is absurd."
my questions are:
(a) : why in first paragraph he said: "any enumeration of $ D $   converges to $x‎_{n‎_{0}‎}‎ $  and hence $ D $   has only one accumulation point, a contradiction." 
(b): why in the last line he said:" Thus $ S $.  can have no accumulation point, since if $ p $.  were such a point, then for some $ k‎ ‎‎\in‎‎ ‎\omega‎ $, $ p‎ =‎ ‎x‎_{k}‎‎ $ and from the construction, we would have that ‎‎$ k‎ ‎‎\geq ‎n‎_{j}‎‎ $  for each $ j‎ ‎‎\in‎‎ ‎\omega‎ $, which is absurd."
(c) : can we say that $ D $   is discreet and closed subset of $ X $.?
plese help me.

 A: (a) Suppose that $D\setminus V$ is finite for each nbhd $V$ of $x_{n_0}$, and let $D=\{y_k:k\in\omega\}$ be any enumeration of $D$; I claim that the sequence $\langle y_k:k\in\omega\rangle$ converges to $x_{n_0}$. To prove this, let $V$ be any open nbhd of $x_{n_0}$; we must show that there is an $m\in\omega$ such that $y_k\in V$ for all $k\ge m$. By hypothesis $D\setminus V$ is finite. If $D\setminus V\ne\varnothing$, let $m=1+\max\{k\in\omega:y_k\in F\}$; otherwise let $m=0$. In either case $y_k\in V$ for every $k\ge m$, and $\langle y_k:k\in\omega\rangle\to x_{n_0}$.
(b) Suppose that $p$ is an accumulation point of $S$. Recall that $X=\{x_k:k\in\omega\}$, and of course $p\in X$, so $p=x_k$ for some $k\in\omega$. Now let $j\in\omega$ be arbitrary; $p$ is an accumulation point of $S\cap D_j$. By construction 
$$n_j=\min\{\ell\in\omega:x_\ell\text{ is an accumulation point of }D_j\}\;,$$
and $k\in\{\ell\in\omega:x_\ell\text{ is an accumulation point of }D_j\}$, so $k\ge n_j$. Since $j$ was arbitrary, we have $k\ge n_j$ for each $j\in\omega$. For each $j$ we have $x_{n_j}\in V_j$ and $D_{j+1}\cap V_j=\varnothing$, and $x_{n_\ell}\in D_{j+1}$ if $\ell>j$, so the points $x_{n_j}$ are distinct: if $j<\ell<\omega$, $x_{n_j}\ne x_{n_\ell}$. This means that $\{x_{n_j}:j\in\omega\}$ is infinite, so $\{n_j:j\in\omega\}$ is infinite, and there is no natural number $k$ such that $k\ge n_j$ for all $j\in\omega$.
(c) No, definitely not. $\Bbb Q$ is a countable $KC$ space with lots of infinite subsets that are neither closed nor discrete.
