Problem about $[x]$ 
$$[x]-2[x/2]\leq 1$$
  Equivalently, $[x]-2[x/2]$ assumes only the values 0 and 1.

It seems easy, but I don't know how to prove it...

Let $f(x)=[x]-2[x/2]$.
Then 
$$f(x+2)= [x+2]-2[(x+2)/2]=[x]+2-2[x/2+1]=[x]+2-2[x/2]-2=f(x) \,.$$
Thus $f$ is periodic with period $2$. It suffices to prove that $f(x) \leq1$ for all $x \in [0,2)$. But this is obvious since then $[x] \leq 1, [x/2] \geq 0$.
