We assert that the real function $x \mapsto \sin\frac{1}{x}$ is not uniformly continuous on the open interval $(0,\,1)$ .

For proving this, we make the antithesis that there exists a positive number $\delta$ such that $$|f(x_1)\!-\!f(x_2)| \;<\; 1 \quad \mbox{always when} \quad x_1,\,x_2 \in (0,\,1)\;\;\mbox{and}\;\; |x_1\!-\!x_2| < \delta.$$ Choose $$x_1 \;=\; \frac{1}{\frac{\pi}{2}\!+\!2n\pi},\;\; x_2 \;=\; \frac{1}{\frac{3\pi}{2}\!+\!2n\pi}$$ where the integer $n$ is so great that $x_1 < \frac{\delta}{2}$ , $x_2 < \frac{\delta}{2}$ . Then we have $$|x_1\!-\!x_2| \;\leqq\; |x_1|+|x_2| \;<\; \delta.$$ However, $$f(x_1)\!-\!f(x_2) \;=\; 1\!-\!(-1) \;=\; 2.$$ This contradictory result shows that the antithesis is wrong.