Show that $x\sin(x)$ is not uniformly continuous on $\Bbb R$ For one of my homework problems, I'm supposed to show that $f(x)=x\sin(x)$ is not uniformly continuous on the real numbers, but this doesn't seem right to me. 
I know that if a function is continuous on a closed and bounded interval then it is uniformly continuous on that interval. I also know that f is continuous on $\Bbb R$. It seems like a simple induction would show that f is UC on $\Bbb R$, so what's going on? 
Please don't actually tell me how to solve the problem, but if you can tell me what I'm missing I would appreciate it. 

We shall assume $x \sin(x)$ is uniformly continuous on $\mathbb{R}$ and reach a contradiction. Let $\varepsilon>0$ be given. Since $x \sin(x)$ is uniformly continuous on $\mathbb{R}$ there is $\delta>0$ so that for all $k \in \mathbb{N}$ we have
\begin{equation} 
\begin{split}
| 2\pi \big(k +\frac{\delta}{2}\big) \sin\big(2\pi \big(k +\frac{\delta}{2}\big)\big) - 2\pi \big(k -\frac{\delta}{2}\big) \sin\big(2\pi \big(k -\frac{\delta}{2}\big)\big)| &= 4 \pi k \sin \big(\frac{\delta}{2}\big) \\
&<\varepsilon.
\end{split}
\end{equation}
However by the Archimedean property, we know that there is a positive integer $\hat{k}$ so that $4 \pi \hat{k} \sin \big(\frac{\delta}{2}\big)> \varepsilon$. This is a contradiction, and so the result is proved.
It should not be hard to see the above equality. Just do a little work to get from the left side to the right side of it.
