I have seen a lot of posts that have solved this already by taking two sequences $(a_n), (b_n)$, then showing $\lim(a_n - b_n) = 0$ but $|f(a_n) - f(b_n)| > \epsilon$, but I would like to show nonuniform continuity with a different approach.
My question is how would you pick two points (not sequences), $x_{\delta}$ and $u_{\delta}$ such that $|x_{\delta} - u_{\delta}| < \delta$ FOR ALL $\delta > 0$?
In particular using this criteria for nonuniform continuity to show that xsinx is not uniform continuous:
ii) There exists an $\epsilon_0 > 0$ such that for every $\delta > 0$ there are points $x_{\delta}, u_{\delta}$ in $A$ such that $|x_{\delta} - u_{\delta}| < \delta$ and $|f(x_{\delta}) - f(u_{\delta})| \geq \epsilon_0$ for all $n \in \mathbb{N}$.
You don't need $$|x_\delta-u_\delta|<\delta$$ for all $\delta>0$. According to the $\varepsilon-\delta$ definition, what you need is to find $\varepsilon_0$ such that for any $\delta$ such a pair of points $x_\delta,u_\delta$ exists, but the values can change with $\delta$.
You may choose for instance $\varepsilon_0=1$. Then for any $\delta>0$ take $x_\delta=2k\pi$ for some integer $k=k(\delta)$ sufficiently large thanks to the oscillation of the cosine there will exist some
$$u_\delta\in(2k\pi-\delta,2k\pi+\delta)$$ such that $$ |u_\delta|\geq1 $$ as desired.
Hope this helps.
PS: the result is not true if you stay within a compact set.