Productive property of normality of a topological space I read in my text book that 

Normality property is not finitely productive.

(i.e if $(X_1, \mathcal T_1)$ , $(X_2, \mathcal T_2)$ be two Topological Spaces being normal,then their topological product may not be normal).
I don't know any counterexample in this context to show that this is the case.
Can anyone help me by giving an example to illustrate this.
Thank you.
 A: Let $S$ denote the Sorgenfrey line; that is, $S=\mathbb{R}$ with the topology generated by the sets of the form $[a,b)$.
It's not too difficult to verify that $S$ is normal.
We argue that the product $S^2$ is not normal. Suppose by contradiction that it is normal. Since $\mathbb{Q}^2$ is a dense subset of $S^2$, two continuous functions $f,g:S^2\to\mathbb{R}$ that agree on $\mathbb{Q}^2$ must be equal.
If $\mathcal{F}:=\{f : S^2\to\mathbb{R} \mid f\ \text{is continuous}\}$, then this observation shows that
$$
|\mathcal{F}| = |\{f:\mathbb{Q}^2\to\mathbb{R} \mid f\ \text{is continuous}\}| \leq |\{f:\mathbb{Q}^2\to\mathbb{R}\}| = |\mathbb{R}^{\mathbb{Q}^2}|=(2^{\aleph_0})^{\aleph_0^2} = 2^{\aleph_0}.
$$
Consider the antidiagonal $D:=\{(x,-x) \mid x\in\mathbb{R}\}$. This is a closed discrete subset of $S^2$ of cardinality $2^{\aleph_0}$. Hence any function $f:D\to\mathbb{R}$ is continuous on $D$, and by the assumption that $S^2$ is normal, we deduce by Tietze's extension theorem that any such function has an extension to a continuous function from $S^2$ into $\mathbb{R}$. This shows that
$$
|\mathcal{F}| \geq |\{f:D\to\mathbb{R}\}|=|\mathbb{R}^D| = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{2^{\aleph_0}}.
$$
Hence $2^{2^{\aleph_0}}\leq|\mathcal{F}|\leq 2^{\aleph_0}$, which is a contradiction because $2^{\aleph_0}<2^{2^{\aleph_0}}$.
A generalization of this idea is found in Jones' Lemma.
A: The Michael line is the following topology on $\mathbb{R}$: If $\mathcal{T}_e$ is the usual topology on $\mathbb{R}$ and $\mathbb{P}$ is the subset of irrationals define a topology $\mathcal{T}_m = \mathcal{T}_e \cup \{A: A \subseteq \mathbb{P}\}$.
This is the smallest topology finer than the usual topology which makes $\mathbb{P}$ an open and discrete subspace.
Then $(\mathbb{R}, \mathcal{T}_m)$ is normal but its product with $\mathbb{P}$ (in the subspace topology from the usual topology of $\mathbb{R}$) is not normal. The irrationals are a metric space so certainly normal.
For full proofs see this post and there we also find a variant of this space which is Lindelöf, such that a product with a metric space is not normal.
There even is a so-called Dowker space $X$ (which is harder to describe) that is normal and such that $X \times [0,1]$ is not normal.
