Proving that $f(x) = \vert x \vert^{\alpha}$ is Holder continuous, inequality help

On any set which is bounded away from $0$, $f$ is actually Lipschitz, as you can check by bounding its derivative. So on such sets $f$ is trivially $\alpha$-Holder continuous, since it is actually $1$-Holder continuous. So now try showing that $f$ is $\alpha$-Holder continuous at $0$.

For $x_0 = x$, $x=0$ or $x_0=0$ the inequality is obvious. W.l.o.g. let $x > x_0$. Then

$$ \frac{\left|x^\alpha-x_0^\alpha\right|}{|x-x_0|^\alpha} =\frac{\left|1-\left(\frac{x_0}{x}\right)^\alpha\right|}{|1-\left(\frac{x_0}{x}\right)|^\alpha} $$

Since $0 < 1- x_0/x < 1$ and $0<\alpha\leq 1$we get $|1-\left(\frac{x_0}{x}\right)|^\alpha \geq |1-\left(\frac{x_0}{x}\right)|$. Similarly $\left|1-\left(\frac{x_0}{x}\right)^\alpha\right| \leq \left|1-\left(\frac{x_0}{x}\right)\right|$. So

$$ \frac{\left|x^\alpha-x_0^\alpha\right|}{|x-x_0|^\alpha} \leq \frac{\left|1-\left(\frac{x_0}{x}\right)\right|}{|1-\left(\frac{x_0}{x}\right)|} = 1 $$