Prove that $ \lfloor 2x \rfloor \leq 2\lfloor x \rfloor + 1$ 
Theorem. For $x \in \mathbb R$, $$2 \lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2 \lfloor x \rfloor +1.$$

I tried to prove this theorem by first proving a few helper theorems. I have proved the following Lemma.

Lemma. For $x \in \mathbb R$ and $n \in \mathbb N$, 
  $$n \leq x \Leftrightarrow n \leq \lfloor x \rfloor. $$

This allowed me to prove the first inequality, because 
$$2 \lfloor x \rfloor = \lfloor x \rfloor  + \lfloor x \rfloor \leq x + x = 2x, $$
and because $2 \lfloor x \rfloor$ is integer, $2 \lfloor x \rfloor \leq \lfloor 2x \rfloor$ due to the Lemma above. However, I can not get a hold of the other inequality. Anyone any ideas? Note: I have seen approaches using the fact that $x = \lfloor x \rfloor + x_1$ with $x_1 \in [0,1)$, however I prefer to not take such an approach, and rather work with the properties described above.

Let $0\leq\{x\}<\frac{1}{2}.$
Hence, $$[2x]=[2[x]+2\{x\}]=2[x].$$
Let $\frac{1}{2}\leq\{x\}<1$.
Hence, $$[2x]=[2[x]+2\{x\}]=2[x]+1$$
and we are done!
