A hereditarily Lindelöf, minimal KC-space is compact A space is said to have the finite derived set property if each infinite subset $A ⊂ X$ contains an infinite subset with only finitely many accumulation points in X.

A hereditarily Lindelöf, minimal KC-space is compact.

Proof: Suppose that $(X, σ)$ is a hereditarily Lindelöf minimal $KC$-space;so 
X has the FDS-property. If (X, σ) is not compact then since it is Lindelöf, it is not
countably compact and hence there is some countably infinite closed discrete subspace
$D =\{d_n : n \in ω \} ⊆ X$. Fix $p \in X$ and a free ultrafilter $G \in βω - ω$ and define a new topology $μ$ on $X$ as follows:
(i) If $p \not\in U$, then $U in μ$ if and only if $U \in σ$,
and
(ii) If $ p \in U$, then $U \in μ$ if and only if $U \in σ$ and $\{n \in ω : d_n ∈ U\} \in G$
. (X, μ) is a $T_1$-space, $μ ⊂ σ$ and for each $B ⊆ X$, $cl_μ(B) ⊆ cl_σ(B) ∪ \{p\}$; since
$(X, σ)$ has the FDS-property, it follows that $(X, μ)$ does as well. We proceed to show that
$(X, μ)$ is a $KC$-space. To this end, suppose to the contrary that $A$ is a non-closed, compact subset of $(X, μ)$.  $p \in cl_μ(A)$ and there are two cases to consider:
(a) If $p \not\in A$, then $μ|A = σ|A$ and so $A$ is compact and hence closed in $(X, σ)$. Thus
$U = X - A$ is open and $p ∈ U$. If $\{n \in ω : dn ∈ A \} \not\in G$, then $\{n \in ω : d_n ∈ D - A \} \in G$
and for each $d ∈ D - A$, $d ∈ U$ and so $p ∈ U ∈ μ$ contradicting the fact that $p ∈ cl_μ(A)$.
Thus $\{n ∈ ω : d_n ∈ A \} ∈ G$ and hence there is some infinite set $S ⊂ A ∩ D$ such that
$\{n ∈ ω : d_n ∈ S\} \not\in G$ and $S$ is then an infinite closed discrete subset of $A$ in $(X, μ)$,
implying that $(A, μ|A)$ is not compact, again a contradiction.
(b) If $p ∈ A$, then $cl_μ(A) = cl_σ(A)$, implying that $A$ is not closed in $(X, σ)$. Thus $A$ is
not compact and since $A$ is Lindelöf, it is not countably compact in $(X, σ)$. Thus there is
a countably infinite, discrete subset $C ⊆ A$ which is closed in $(A, σ|A)$. However, $C$ is not
closed in $(A, μ|A)$ and so $cl_μ(C)∩A = C∪ \{p\}$. This implies that $\{n ∈ ω : d_n ∈ cl_μ(C) \} ∈ G$.
If $P = \{n ∈ ω : d_n ∈ C \}$ is infinite, then there is some infinite subset $S ⊆ P$ such that
$S \not\in G$ and hence $ \{d_n : n ∈ S\}$ is a closed, discrete subspace of $(A, μ|A)$, contradicting
the compactness of this space. If, on the other hand, $P$ is finite, then since $(X, μ)$ has the
$FDS$-property, there is an infinite subset $B ⊆ C$ with only a finite number of accumulation
points in $(X, μ)$. Thus $\{n ∈ ω : d_n ∈ cl_μ(B)\} \not\in G$ which implies that $B$ is closed and
discrete in $(A, μ|A)$, implying in its turn that $A$ is not compact in $(X, μ)$.

I would like to know:
1: Why $p \in cl_μ(A)$ ?
2: Why 
there is some infinite set S ⊂ A ∩ D such that
  $\{n ∈ ω : d_n ∈ S\} \not\in G$ ? And $S$ is then an infinite closed  and discrete subset of $A$ in $(X, μ)$,
  implying that $(A, μ|A)$ is not compact?
In part (b)
3: Why
$cl_μ(C)∩A = C∪ \{p\}$?why can we say $\{n ∈ ω : d_n ∈ cl_μ(C) \} ∈ G$, I mean, why $d_n \in  cl_μ(C)$ not $C$?and in the last line  $\{n ∈ ω : d_n ∈ cl_μ(B)\} \not\in G$?( why in cl(B)?

 A: $\newcommand{\cl}{\operatorname{cl}}$I’m going to split your (3) into three parts.


*

*Suppose that $p\notin\cl_\mu A$. Let $\mathscr{U}$ be a $\sigma$-open cover of $A$; then $\mathscr{U}'=\{U\setminus\{p\}:U\in\mathscr{U}\}$ is also a $\sigma$-open cover of $A$. Moreover, $\mathscr{U}'\subseteq\mu$, and $A$ is $\mu$-compact, so there is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ such that the finite subset $\{U\setminus\{p\}:U\in\mathscr{U}_0\}$ of $\mathscr{U}'$ covers $A$. Clearly $\mathscr{U}_0$ is a finite subset of $\mathscr{U}$ that covers $A$ so $A$ is $\sigma$-compact and therefore $\sigma$-closed, since $\langle X,\sigma\rangle$ is $KC$. Finally, $\cl_\mu A\subseteq\cl_\sigma A\cup\{p\}=A\cup\{p\}$, and by hypothesis $p\notin\cl_\mu A$, so $\cl_\mu A=A$, contradicting the hypothesis that $A$ is not closed in $\langle X,\mu\rangle$.

*Let $N_A=\{n\in\omega:d_n\in A\}$; at this point in the argument we know that $N_A\in G$. It’s a basic fact about ultrafilters that if $G$ is an ultrafilter, and $T_0\cup T_1\in G$, then exactly one of $T_0$ and $T_1$ belongs to $G$. $N_A\in G$, and $G$ is a free ultrafilter, so $N_A$ is infinite; partition $N_A$ into two infinite sets $T_0$ and $T_1$. Then exactly one of $T_0$ and $T_1$ does not belong to $G$, say $T_0$. Let $S=\{d_n\in D:n\in T_0\}$; then $S$ is infinite (since $T_0$ is), $S\subseteq A\cap D$, and $T_0=\{n\in\omega:d_n\in S\}\notin G$. Let $V=X\setminus S$; $S$ is closed and discrete in $\langle X,\sigma\rangle$ (since $D$ is), so $V\in\sigma$. And $\{n\in\omega:d_n\in V\}\supseteq T_1\in G$, so $U\in\mu$ as well, and therefore $S$ is $\mu$-closed. The same argument shows that every subsets of $S$ is $\mu$-closed and hence that $S$ is discrete in $\langle X,\mu\rangle$.

*$C$ is $\sigma$-closed but not $\mu$-closed in $A$, so there is at least one point of $A$ that belongs to $\cl_\mu C\setminus\cl_\sigma C$. On the other hand, the only point of $X$ at which the topologies $\sigma$ and $\mu$ differ is $p$, so $p$ is the only point of $A$ that could possible be in $\cl_\mu C\setminus\cl_\sigma C$. It must therefore be the case that $A\cap(\cl_\mu C\setminus\cl_\sigma C)=\{p\}$ and hence that $$A\cap\cl_\mu C=A\cap(\{p\}\cup\cl_\sigma C)=A\cap(\{p\}\cup C)=\{p\}\cup C\;.$$ 

*$C$ might be disjoint from $D$, so we certainly can’t expect that $\{n:d_n\in C\}\in G$. Let $D_C=D\cap\cl_\mu C$, let $N_C=\{n\in\omega:d_n\in D_C\}$, and suppose that $N_C\notin G$. Let $V=X\setminus D_C$; $D_C$ is $\sigma$-closed in $X$, so $V\in\sigma$. Moreover, $\{n\in\omega:d_n\in V\}=\omega\setminus N_C$, and $N_C\notin G$, so $\{n\in\omega:d_n\in V\}\in G$. 

*At this point we are assuming that $P$ is finite and hence that $C\cap D$ is finite, and we are also assuming that $B$ is an infinite subset of $C$ with only finitely many $\mu$-accumulation points. Clearly $B\cap D\subseteq C\cap D$ is finite, and $B\setminus\cl_\mu B$ is finite, so $D\cap\cl_\mu B$ is finite. Thus, $\{n\in\omega:d_n\in\cl_mu B\}$ is finite. Finally, $G$ is a free ultrafilter, so no finite subset of $\omega$ belongs to $G$, and therefore $\{n\in\omega:d_n\in\cl_\mu B\}\notin G$.
