Proof of identity $\sqrt {xy} = \sqrt x \sqrt y$ for $x,y \in \mathbb R^+$
Hint: If you have proved that $\sqrt{x}\sqrt{y}$ is a non-negative number such that $(\sqrt{x}\sqrt{y})^2 = xy$, then this means by definition that $\sqrt{x}\sqrt{y} = \sqrt{xy}$.
This is based on user119191's comment:
From definition, let $a,b\in \mathbb R^+ :a=\sqrt {xy}, b=\sqrt x \sqrt y$
$a^2=xy,b^2=xy\rightarrow a^2=b^2\rightarrow a=b \to\sqrt {xy}=\sqrt x \sqrt y$