Proving that $f(x)= \sin (2\pi x [ x])$ is not uniformly continuous on $\mathbb{R}$

Here, $[\cdot ]$ denotes the greatest integer function. I tried a lot but couldn't succeed. Also, I noticed that $f$ is bounded and continuous. To prove it is not uniformly continuous on $\mathbb{R}$, we assume the contrary. Now, we let $\varepsilon = 1$. Thus, there is $\delta > 0$ such that if $x,y \in \mathbb{R}$ with $\left| x-y \right| < \delta$, then we have $\left| \sin (2\pi x [ x]) - \sin (2\pi y [ y])\right| < 1$.

I was trying to consider the values of $x$ where $f(x)=\pm 1$. But it certainly didn't help. Hints would be appreciated.

Set $$\epsilon=0.5$$ and let $$\delta>0$$.Choose a natural number $$N$$ such that $$1/N<\delta$$. Then the points $$x=N$$, and $$y=N+\frac{1}{4N}$$ satisfy $$|x-y|<\delta$$ and $$|f(x)-f(y)|>0.5$$.