Let $x\in \mathbb{R}$. Show that $2⌊x⌋ ≤ ⌊2x⌋ ≤ 2⌊x⌋+1$

Question

So far I got the below and is unsure where to go from there...

By definition, $\left\lfloor x\right\rfloor =m$ such that

$$m\leq x< m+1\text{ }\forall x\in\mathbb{R}\wedge \forall m\in\mathbb{Z}$$ $$\iff m+n\leq x+n<m+n+1\text{ }\forall n\in\mathbb{Z}$$ $$\therefore\left\lfloor x+n\right\rfloor = m+n=\left\lfloor x\right\rfloor+n$$ $$\forall x\in\mathbb{R},\exists\left\lfloor x\right\rfloor\in\mathbb{Z}:\left\lfloor x\right\rfloor\leq x < \left\lfloor x\right\rfloor+1$$

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Edit after receiving hints:

Let $⌊x⌋=m$ with $m\in \mathbb{Z}$. Then there exists $\varepsilon \in [0,1)$ such that $x=m+\varepsilon$. Thus, we have: $2⌊x⌋=2m ≤ ⌊2x⌋=⌊2m+2\varepsilon⌋=2m+⌊2\varepsilon⌋$, which proves the first inequality.

2⌊x⌋+1= 2m+1, and ⌊2ε⌋≤1 for ε∈[0,1). Thus,⌊2x⌋= 2m+⌊2ε⌋≤2⌊x⌋+1= 2m+1, which proves the second inequality.

Answer 1

Hint: Show that $\lfloor 2m+2\varepsilon\rfloor=2m+\lfloor2\varepsilon\rfloor$. What values can $\lfloor2\varepsilon\rfloor$ take?