Uniform Continuity of $x \sin x$

Let $f(x)=x\sin x$ and $x_n=\pi n$ and $y_n=\pi n+\frac{1}{n}$ then $\displaystyle\lim_{n\to\infty}( y_n-x_n)=0$ and

$f(y_n)-f(x_n)=(\pi n+\frac{1}{n})\sin(\pi n+\frac{1}{n})-\pi n\sin(n\pi)=_{n\longrightarrow\infty} \pi(-1)^n+o(1)\not \to 0$

Elaborating on David Mitra's Comment:

From continuity of $x\sin(x)$, you have that $$\forall \epsilon >0, \forall y, \exists \delta: |x-y|<\delta\implies |x\sin(x)-y\sin(y)|<\epsilon$$

But define $$p=x+2n\pi;\;\;\;q=p=y+2n\pi$$

Then you have $|p-q|<\delta$ but $$|(x+2n\pi)\sin(x+2n\pi)-(y+2n\pi)\sin(y+2n\pi)|$$ $$=|(x\sin(x)-y\sin(y))+2n\pi(\sin(x)-\sin(y))|>\epsilon$$ for sufficiently large $n$. An undesirable result for uniform continuity.

Since $x\sin(x)$ is continuous, we won't be able to show discontinuity. It is the uniformity of the continuity that we have to consider. $f$ is uniform continuous if and only if

$$\forall\epsilon\gt0,\exists\,\delta\gt0:\forall x,y\in\mathbb{R},|x-y|\le\delta\implies|f(x)-f(y)|\le\epsilon\tag{1}$$

The inverse of $(1)$ is

$$\exists\,\epsilon\gt0:\forall\,\delta\gt0,\exists\,x,y\in\mathbb{R}:|x-y|\le\delta\wedge\,|f(x)-f(y)|\gt\epsilon\tag{2}$$

We can take $\epsilon=1$. Given $\delta>0$, let $\eta=\min(\delta,\pi/2)$ and $x=2\pi\left\lceil\dfrac1{2\pi\sin(\eta)}\right\rceil$ and $y=x+\eta$. Then $f(x)=0$ and $$ \begin{align} f(y) &=(x+\eta)\sin(x+\eta)\\ &=(x+\eta)\sin(\eta)\\ &\gt x\sin(\eta)\\ &\ge1 \end{align} $$