Does "regular" implies collectionwise hausdorff? Does "regular"  implies collectionwise hausdorff?

A topological space is said to be collectionwise Hausdorff if given any closed discrete collection of points in the topological space, there are pairwise disjoint open sets containing the points. 

I believe it and try to proof it, however I'm not sure. Thanks in advance:)
 A: No, it doesn't. The Sorgenfrey plane is regular, as the Sorgenfrey line is normal, but it isn't collectionwise Hausdorff as $\{(-x,x)\mid x \in \mathbb R\}$ is discrete and closed, but there aren't pairwise disjoint open neighbourhoods. 
To prove this, let $U_x = [-x,-x+\epsilon_x) \times [x,x+\epsilon_x)$ be a neighbourhood of $(-x,x)$ in the Sorgenfrey plane, then there is some $\epsilon > 0$ such that $E_\epsilon= \{x \in \mathbb R\mid \epsilon_x > \epsilon \}$ is uncountable (as $\mathbb R = \bigcup_n E_{1/n}$). Then $E_\epsilon$ has an accumulation point in $\mathbb R$, choose $x_1, x_2 \in E_\epsilon$ with $|x_1 - x_2| < \frac \epsilon 4$, then $U_{x_1} \cap U_{x_2}\ne \emptyset$.

To answer your addional question in the comment below: Let $X$ be regular and $A =\{a_n \mid n \in \mathbb N\}$ closed, discrete and countable. For each $n \in \mathbb N$ there are by regularity and closedness of all subsets of $A$ in $X$ disjoint open sets $U_n \ni x_n$, $V_n \supseteq A \setminus \{x_n\}$. Now let $W_n = U_n \cap \bigcap_{i < n} V_i$, then $W_n$ is an open set containing $x_n$ (as finite intersection of such). Now let $n < m$, then $W_n \subseteq U_n$, $W_m \subseteq V_n$, so $W_n \cap W_m = \emptyset$. Hence $X$ is (don't know if this is a common term, but IMHO it describes the property) countable collectionwise Hausdorff. 
