How to show that the product of two irrational numbers may be irrational?

Well, if all you know is that $\sqrt{2}$ is irrational, try the pair of $\sqrt{2}$ and $\sqrt{2}+1$ - both of which are clearly irrational, and their product is $2+\sqrt{2}$, which is also clearly irrational. Then we don't have to know anything other than that $\sqrt{2}$ is irrational and an irrational plus a rational is still irrational.

What about $\sqrt{2}\times \sqrt{3}$?

Another way to tackle this is to prove that if $n$ is irrational, so is $\sqrt{n}$. (This is straightforward from the definition of rationality.) Then it's easy to see that for irrational $n$, $$\sqrt{n} \cdot \sqrt{n} = n$$ is an irrational product of irrational numbers.