GIVE AN EXAMPLE :
X:normal
Y:normal
BUT X X Y : not normal
normal is not productive
The classic example (given in Kelley's General Topology, and attributed by him to Dieudonné and Morse, independently) uses ordinal spaces. Let $\displaystyle \Omega_0$ be the set of all ordinal numbers less than the first uncountable ordinal $\displaystyle \Omega$ and let $\displaystyle \Omega' = \Omega_0\cup\{\Omega\}$, each with the order topology. Then $\displaystyle \Omega_0$ and $\displaystyle \Omega'$ are both normal, but $\displaystyle \Omega_0\times\Omega'$ is not. (Let $\displaystyle A = \{(x,x):x\in\Omega_0\}$ and $\displaystyle B = \Omega_0\times\{\Omega\}$. Then A and B are closed and disjoint in $\displaystyle \Omega_0\times\Omega'$ but do not have disjoint open neighbourhoods. The proof is not that easy. You would do best to look it up in Kelley's book.)
Another example is a Sorgenfrey plane $\displaystyle \mathbb{R}_l^{2}$.
Even though a Sorgenfrey line is normal, the Sorgenfrey plane is not normal.
Let $\displaystyle \Delta = \{(x, -x) | x \in R \}$. As shown by the above link, no disjoint open sets can separate the disjoint closed sets $\displaystyle K= \{(x, -x) | x \in \mathbb{Q} \}$ and $\displaystyle \Delta \setminus K$ in $\displaystyle \mathbb{R}_l^{2}$. Intuitively, unions of the open rectangles containing $\displaystyle K$ and unions of the open rectangles containing $\displaystyle \Delta \setminus K$ in the above link somehow overlaps all the time. A rigorous proof needs more elaboration.
In contrast, $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{R} \setminus \mathbb{Q}$ are disjoint closed sets in a Sorgenfrey line $\displaystyle \mathbb{R}_l$, which can be separated by disjoint open sets in $\displaystyle \mathbb{R}_l $ (they are both clopen sets).