If $f$ is a function, by definition, if $a=b$ then $f(a)=f(b)$. The floor function is an example of a function and so you have $a=b$ implies $\lfloor a\rfloor = \lfloor b\rfloor$.
Now... some functions will retain inequalities too. If a function is monotonic increasing that means by definition that if $a\leq b$ then $f(a)\leq f(b)$. The floor function is an example of a monotonic increasing function. That is to say, if $a\leq b$ then it is true that $\lfloor a\rfloor \leq \lfloor b\rfloor$.
Not every function is monotonic increasing however. Take for example $f(x)=x^2$. You have $-3\leq 1$ but $(-3)^2\not\leq (1)^2$.
Finally, some functions will retain strict inequalities as well. If a function is strictly monotonic increasing that means by definition that if $a<b$ then $f(a)<f(b)$. The floor function is not strictly monotonic, for example how $1.2<1.5$ but $\lfloor 1.2\rfloor \not\lt \lfloor 1.5\rfloor$ since they both result in the same value of $1$.
As for a hint on how to prove that $a\leq b\implies \lfloor a\rfloor \leq \lfloor b\rfloor$, rewrite $b$ as $a+(b-a)$ and recognize that $(b-a)\geq 0$. Further, apply the definition of the floor function as being the largest integer less than or equal to its input.
Let $x\geq 0$. Then let $c = \lfloor a\rfloor$ be the largest integer less than or equal to $a$. It follows that $c$ is an integer which is less than or equal to $a+x$, whether it is the largest one or not... and so whatever the largest integer less than or equal to $a+x$ is will be at least as large as $c$. As such we have $\lfloor a\rfloor \leq \lfloor a+x\rfloor$, showing that the floor function is monotonically increasing.