If $x- \lfloor x\rfloor \geq \frac 12$, then $ x = \lfloor x \rfloor + \epsilon$ where $\frac 12 \le \epsilon < 1 \implies 1 \le 2 \epsilon < 2 \implies \lfloor 2\epsilon \rfloor = 1$
So \begin{align} \lfloor 2x \rfloor &= \lfloor 2\lfloor x \rfloor + 2 \epsilon \rfloor \\ &= 2\lfloor x \rfloor + \lfloor 2 \epsilon \rfloor & \text{(If $n$ is an integer, then $\lfloor n + \xi \rfloor = n + \lfloor \xi \rfloor$)}\\ &=2\lfloor x\rfloor+1 \\ \end{align}
Added 12/25/2017
If $x- \lfloor x\rfloor \geq \frac{1}{2}$
Then $2x- 2\lfloor x\rfloor \geq 1$
So $2\lfloor x \rfloor + 1 \le 2x$
Since $ x - \lfloor x \rfloor < 1$
Then $2x \le 2\lfloor x \rfloor + 2$
So $2\lfloor x \rfloor + 1 \le 2x < 2\lfloor x \rfloor + 2$
Since $2\lfloor x \rfloor + 1$ is an integer, then $2\lfloor x \rfloor + 1 \le \lfloor 2x \rfloor$
Since $2\lfloor x \rfloor + 2$ is an integer, then $\lfloor 2x \rfloor < 2\lfloor x \rfloor + 2$
So $(2\lfloor x \rfloor + 1) \le \lfloor 2x \rfloor < (2\lfloor x \rfloor + 1) + 1$
By the definition of the floor function, then $2\lfloor x \rfloor + 1 = \lfloor 2x \rfloor$