analysis - Prove that $f(x) = x^{1/5}$ is continuous everywhere

Need to prove that $f(x) = x^{1/5}$ is continuous everywhere, where $f: \mathbb{R} \to \mathbb{R}$:

from definition we need to show that given $ \epsilon > 0 $ $\exists \delta > 0 $ s.t. $|x-x_0|<\delta \Rightarrow \left|x^{\frac{1}{5}} - x_0^{\frac{1}{5}}\right| < \epsilon$ for any point $x_0 \in \mathbb{R}$

I have a proof but it's somewhat unjustified:

consider
$\left|x^\frac15 - x_0^\frac15\right| \geq \left|x^\frac15\right| - \left|x_0^\frac15\right| $ from the triangle inequality since $\left|x^\frac15\right| < |x|$ and $\left|x_0^\frac15\right| < |x_0|$ then $\left|x^\frac15 - x_0^\frac15\right| \geq \left|x^\frac15\right| - \left|x_0^\frac15\right| < |x| - \left|x_0\right| = \left|x-x_0\right| = \delta$ so we can choose $\delta = \epsilon$?

Overall I'm not happy with the proof, in the last inequality I don't think I can just state that delta = epsilon and be done, but I have no idea what else to do. I also am not sure about this step $|x| - |x_0| = |x-x_0|$and $\left|x^\frac15\right| < |x|$ and $\left|x_0^\frac15\right| < |x_0|$ that step also...

if anyone could help me out.. thank you