How to show that $x^{1/n}$ is uniformly continuous

Suppose $x \ge y$ and $\eta_n := x^{1/n} - y^{1/n}$. Then $\eta_n \ge 0$ and $$x = (y^{1/n} + \eta_n)^n \ge (y^{1/n})^n + \eta_n^n = y + \eta_n^n$$ Thus $0 \le \eta_n \le (x - y)^{1/n}$. By symmetry it follows that $|x^{1/n} - y^{1/n}| \le |x - y|^{1/n}$. Therefore, if $|x - y| < \epsilon^n$ then $|x^{1/n} - y^{1/n}| < (\epsilon^n)^{1/n} = \epsilon$.

The point is that the rate of increase is the fastest at $x_0=0$, and in that case you have $|f(x)-f(0)|=x^{1/n}<\epsilon$ if $x<\epsilon^n$. So what you want to show is that $|f(x+h)-f(x)| \leq |f(h)-f(0)|$ if $x>0$ and $h>0$. This ultimately follows from the fact that the function is concave down.