Suppose I want to show that
$$ a\times b\leq c\times d $$ where $a,b,c,d\in [0,1]$ and $a\leq c$.
Now, we can see that
$$ a\times b\leq c\times d \hspace{0.5cm}\Leftrightarrow \hspace{0.5cm}b\leq \underbrace{\overbrace{\frac{c}{a}}^{\geq 1}\times d}_{\geq d} $$
Therefore: is it correct to say that in order to show $a\times b\leq c\times d$ it is sufficient (but not necessary) showing that $b\leq d$?
If $0\leq a\leq c$ and $0\leq b\leq d$ then indeed we have $ab\leq cd$ but your reasoning is not sound.
If $a=0$ then $\frac{c}{a}$ is not defined, so it must be treated as a special case.