Regular topological spaces need not to be normal I was looking for a counterexample for the following statement:
"A regular topological space need not to be normal."
I don't understand how to use the lemma to prove Theorem 7: http://fac.hsu.edu/worth/topology/separate.html

Lemma. Suppose $X$ is a topological space that is normal and has a countable subset whose closure is the whole space. If $E$ is an uncountable subset of $X$, then $E$ has a limit point.
Theorem 7. $T_3$ does not imply $T_4$ and regular does not imply normal.
Proof. Let $X$ be the real line. Let $B = \{[a, b)\}$. $B$ is a basis for a topology for $X$. The lemma can be used to show that, though it is regular, $X \times X$ is not normal.

Thank you!
 A: The lemma in question says that if $X$ is a separable, normal space, and $E\subseteq X$ is uncountable, then $E$ has a limit point. Since a subset of $X$ has no limit point if and only if it is closed and discrete, this amounts to saying that every closed, discrete subset of a separable, normal space is countable. Unfortunately, this is not quite correct. What is true is that if $C$ is a closed, discrete subset of a separable, normal space, then $|\wp(C)|\le 2^\omega=\mathfrak{c}$. This is Jones’s lemma, and you can find a proof here. (The version that you found is equivalent to Jones’s lemma only if the continuum hypothesis  ($\mathsf{CH}$) holds;  $\mathsf{CH}$ is consistent with the usual axioms of set theory, but so is its negation.)
Now let $X$ be the space described in the partial proof of Theorem $7$; this space is known as the Sorgenfrey line or reals with the lower limit topology. Let $D=\{\langle x,-x\rangle:x\in X\}$, the ‘reverse diagonal’ in the plane. It’s easy to check that $D$ is closed and discrete in $X\times X$. (E.g., for discreteness note that $[x,x+1)\times[-x,-x+1)$ is an open nbhd of $\langle x,-x\rangle$ that contains no other point of $D$.) Clearly $|D|=|\Bbb R|=2^\omega$, so by Cantor’s theorem $|\wp(D)|>2^\omega$. 
$\Bbb Q\times\Bbb Q$ is a countable dense subset of $X\times X$ in the product Sorgenfrey topology, since every non-empty open set in $X\times X$ contains a non-empty set that is open in the Euclidean topology on $\Bbb R\times\Bbb R$. Thus, $X\times X$ is separable, and by Jones’s lemma it cannot be normal. My answer to this question indicates why $X\times X$ is regular (indeed, Tikhonov).
