Uniform Continuity of a function with the simplest way I'm trying to show that 
$f(x) =
\begin{cases}
x \sin\left(\frac1x\right),\quad\text{if $x \in (0,1]$ }\\[2ex]
0, \quad \quad \quad \quad \ \text{if $x=0$}
\end{cases}$ 
is uniformly continuous on $[0,1]$
Let $\epsilon \gt 0$ and let $x, y \in (0,1)$.
Then
$$
 \left|x\sin\frac{1}{x} - y \sin\frac{1}{y} \right|=\left| x\sin\frac{1}{x} - y\sin\frac{1}{x} + y\sin\frac{1}{x} - y \sin\frac{1}{y} \right|
= \left| (x-y)\sin\frac{1}{x} + y \left(\sin\frac{1}{x} - \sin\frac1y\right) \right| $$
and by triangle inequality
\begin{equation}
\label{eq: star}
\left | x\sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x - y| + y \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |
\end{equation}
$$\left |  \sin\frac{1}{x} -\sin\frac{1}{y} \right|= \left | 2 \cos \left ( \frac{1}{2} \left ( \frac{1}{x} + \frac{1}{y} \right ) \right ) \sin \left ( \frac{1}{2} \left ( \frac{1}{x} - \frac{1}{y} \right ) \right ) \right |$$
I could not continue from there. Is there any basic way for this? I've seen other answers on this question but I'm looking for really simple one for formal proof.
Thanks for any help.

Here is an ad-hoc proof. Let $\epsilon>0$ be fixed. We can and do adjust it to $1$ if it is bigger.
The function $|f'|$ is bounded and continuous on $[\epsilon/3, \; 1]$, let $M> 3$ be an upper bound.
We set $\delta = \epsilon/M<\epsilon/3$.
Let $x,y$ be two points in $[0,1]$ at distance $<\delta$, and we can and do assume $0\le x\le y\le x+\delta$. Two cases:


*

*If $x$ is in the interval $[0,\epsilon/3]$ then we have:


$$|f(x)-f(y)|\le |f(x)|+|f(y)|\le x+y\le 2x+(y-x)<
\frac 23\epsilon +\delta<\epsilon\ .
$$


*

*If $x>\epsilon/3$ then both $x,y$ live in the interval $[\epsilon/3,\; 1]$ where we can find an intermediate point $\xi$ with


$$|f(x)-f(y)|=|f'(\xi)|\;|x-y|<M|x-y|<M\delta<\epsilon\ .
$$
