prove that $\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$

First note that $f:\mathbb R\rightarrow \mathbb Z$ given by $f(x)=\lfloor x\rfloor$ is an increasing function that is the identity on the integers. Then note that for positive $x$ we have $0\leq f(x)\leq x$. With this we get $$ f(x)f(y)\leq xy $$ and applying the increasing function $f$ on both sides above noting that the left hand side is an integer we then get: $$ f(x)f(y)=f(f(x)f(y))\leq f(xy) $$ which proves the claim.

Assuming that you mean $\lfloor x \rfloor \lfloor \color{red}{y} \rfloor \le \lfloor xy \rfloor$, write

$$x := a + b, \quad y = c + d, \quad a,c \in \mathbb{N} \cup \{0\}, \quad b,d \in [0,1\rangle.$$

Then $\lfloor x \rfloor \lfloor y \rfloor = ac$ and

$$\lfloor xy \rfloor = \lfloor (a+b)(c+d) \rfloor = \dots$$

HINT: For this you need only a little more than the fact that if $x\ge 0$, then $0\le\lfloor x\rfloor\le x$ and basic facts about manipulating inequalities. Specifically, you need to realize that if $z\in\Bbb R$, $n\in\Bbb Z$, and $n\le z$, then $n\le\lfloor z\rfloor$.