A question of a A. Dow's 2002 paper As the title explains, I would like to understand the Example(4.3)(4), Page 195 of the paper. However, the last part I cannot pass which said: Let us finally prove that $X$ is not weakly discretely generated. 

A space is called weakly discretely generated if for every $A\subset X$ with $cl(A)\not=A$ there is a discrete $D\subset A$ such that $cl(D)\setminus A \not=\emptyset$.

Here is the paper link. Thanks for any help.
 A: $\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}C$ is a dense subset of $\beta\omega\setminus\omega$ of cardinality $2^\omega$. $C$ is dense-in-itself, so its relative topology can be refined to a maximal dense-in-itself topology $\mu$. $X$ is $\langle\beta\omega\setminus\omega,\nu\rangle$, where $\nu$ is the topology generated by the subbase $\tau\cup\mu$, where $\tau$ is the usual topology on $\beta\omega\setminus\omega$.
Now let $x\in C$, and suppose that $A\subseteq X\setminus\{x\}$ with $x\in\cl_\nu A$. $C\in\mu\subseteq\nu$, so $C$ is an open nbhd of $x$ in $X$. Let $A_0=C\cap A$, and let $U$ be any open nbhd of $x$; then $U\cap C$ is an open nbhd of $x$, so $$U\cap A_0=(U\cap C)\cap A\ne\varnothing\;,$$ and therefore $x\in\cl_\nu A_0$. But the relative topology on $C$ as a subspace of $X$ is $\mu$, so $\cl_\nu A_0=\cl_\mu A_0$, and $x\in\cl_\mu A_0$.
Now suppose that $A$ is discrete in $X$; clearly $A_0$ is discrete in $C$. Suppose that $y$ is an isolated point of $C\setminus A_0$; then $y$ has an open nbhd $U\subseteq\{y\}\cup A_0$. But this is impossible: clearly $U$ cannot be dense-in-itself, but $C$ is dense-in-itself, so every non-empty open subset of $C$ is dense-in-itself. Thus, $C\setminus A_0$ is dense-in-itself. Let 
$$\begin{align*}
\mu'&=\left\{U\cup\big(V\cap(C\setminus A_0)\big):U,V\in\mu\right\}\\
&=\left\{U\cup(V\setminus A_0):U,V\in\mu\right\}\;;
\end{align*}$$
then $\mu'$ is a topology on $C$, $\mu'\supseteq\mu$, and we just showed that $\langle C,\mu'\rangle$ has no isolated points. By the maximality of $\mu$ we must have $\mu'=\mu$ and hence $V\setminus A_0\in\mu$, i.e., $A_0$ is closed in $C$. This contradicts the hypothesis that $x\in(\cl_\mu A_0)\setminus A_0$, thereby showing that $A$ cannot be discrete in $X$ and hence that $X$ is not weakly discretely generated. (Specifically, for any $x\in X$ the set $A=X\setminus\{x\}$ fails the condition of Definition 3.2: there is no discrete $D\subseteq A$ such that $(\cl D)\setminus A\ne\varnothing$.)
