<s>[INST] Prove $f(x) = x^2\sin\left(\frac{1}{x}\right)$ is Lipschitz (no use of derivative) 
Prove that $f:\mathbb{R}\to \mathbb{R}$ such that
  $$ f(x) = \left\{ 
  \begin{array}{c l}
    x^2\, \sin\left(\frac{1}{x}\right) & ,\quad x\neq 0\\
    0 & ,\quad x=0
  \end{array} \right.$$
  is Lipschitz (without use of derivatives).

Attempt. I am aware (Lipschitz-continuous $f(x)=x^2\cdot \sin\left(\frac{1}{x}\right)$)
 that: $$|f(x)-f(y)|\leq 3|x-y| ~~~\forall~x,~y\in \mathbb{R},$$ 
but I am looking for a proof, without use of derivatives. 
I tried: for $x,y\neq 0$:
\begin{eqnarray}
x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y}
 &=& (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right ),\nonumber
\end{eqnarray}
so: $$\left | x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right |
\leq |x^2 - y^2| + y^2 \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |.$$
Since:
$$\left | \sin\frac{1}{x} - \sin\frac{1}{y} \right | \leq  \left| \frac{1}{x} - \frac{1}{y}\right |= \frac{|x - y|}{xy},$$
we get:
$$\left | x^2 \sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right |
\leq \left(x+y+\frac{y}{x}\right)|x-y|.$$
Unfortunatelly , the quantity $x+y+\frac{y}{x}$ grows to 
$+\infty$, either for large $x$, or for $x\approx 0.$
Thanks in advance for the help.
 [/INST] Writing
$$
f(x) - f(y) = (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| \le |x|$:
$$
\left| (x^2- y^2) \sin \frac 1x \right| \le |x-y| |x+y|| \frac 1x| \le 2|x-y|
$$
and 
$$
\left|y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right )\right| \le |y^2| 
\left| \frac 1x - \frac 1y \right| = \left|\frac yx\right||x-y| \le |x-y|
$$
and therefore
$$
|f(x)-f(y)|\leq 3|x-y| \, .
$$
For $0 < |x| \le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.
Finally, for $x \ne 0 = y$
$$
|f(x) - f(0)| = \left |x^2 \sin \frac 1x \right| \le |x| = |x-0| \, .
$$
</s>