Prove $x\sin2x$ is uniformly continuous on $(0,1)$. My attempt: 
Take $\delta=?$ Then we have
$$\vert x-y\vert\Rightarrow\vert x\sin2x-y\sin2y\vert$$
$$=\vert x(\sin2x-\sin2y)+\sin2y(x-y)\vert$$
$$\leq\vert x\vert\vert\sin2x-\sin2y\vert+\vert\sin2y\vert\vert x-y\vert$$
$$\leq2\vert x\vert+\vert x-y\vert\leq2+\vert x-y\vert$$
My problem is that now I don't know how to define $\delta$ so that $2+\vert x-y\vert<\epsilon$. I thought of $\delta+2<\epsilon$, but then $\delta<0$ for some $\epsilon>0$
 A: $2+|x-y|$ will never be less than $\varepsilon$, if for example $\varepsilon<1$. Instead, as you wrote, $$|x||\sin 2x-\sin 2y|+|\sin 2y||x-y|\leq$$$$|x||2x-2y|+|x-y|\leq 2|x-y|+|x-y|\leq 3|x-y|,$$ where we use that $|\sin x-\sin y|\leq|x-y|$ for all $x,y\in\mathbb R$.
A: $x\sin2x$ is continuous on $[0,1]$, thus uniformly continuous there, which stays true under the restriction of the domain to $(0,1)$.

To stay with your computation, and stay elementary,
$$
\sin 2x - \sin 2y = 2\sin(x-y)\cos(x+y)
$$
by trigonometric theorems, and thus
$$
|\sin 2x - \sin 2y|\le 2\min(1,|x-y|)
$$
A: You need a tighter upper bound for $|\sin 2x - \sin 2y|$. Note that $\sin 2x - \sin 2y$ is the imaginary part of $e^{i2x} - e^{i2y}$, so $|\sin 2x - \sin 2y| \leq |e^{i2x} - e^{i2y}| \leq |2x - 2y|$, where the last inequality is proved here.
A: Using mean-value theorem is another way.
There is some $M > 0$ such that we have
$0 < x < y < 1$ only if
$$
|x \sin 2x - y \sin 2y| \leq |x-y|\sup_{x < t < y}| \sin 2t + 2t\cos 2t| < M|x-y|;
$$
given any $\varepsilon > 0$, we have $M|x-y| < \varepsilon$ if $|x-y| < \varepsilon/M$; taking $\delta := \varepsilon/M$ suffices.
