How to prove $x^{n}$ is not uniformly continuous

You want to show that there exists some $\epsilon>0$ such that there is no $\delta>0$ for which $$|x-y|<\delta\implies |x^n-y^n|<\epsilon.$$ Try $\epsilon=1$. Thus you want to show that for any $\delta>0$, there is some pair $x,y\in [0,\infty)$ such that $|x-y|<\delta$ yet $|x^n-y^n|\geq 1$. Let's try letting $x=y+\delta/2$. Then we have $|x-y|=\delta/2<\delta$ and $$|x^n-y^n|=(y+\delta/2)^n-y^n\geq (y+\delta/2)y^{n-1}-y^n=\delta/2\cdot y^{n-1}.$$ Can you find some $y$ such that $\delta/2\cdot y^{n-1}$ is at least $1$?

We note that since we have freedom to choose an $\varepsilon>0$, we let $\varepsilon=1$. Therefore, we want to show that for some $x,y$ that $|x^n-y^n|<1$ does not arise from $|x-y|<\delta$. For any $\delta>0$,
if we let $x=\delta$ and $y=\frac{1}{2\delta}$, then $$|x-y|=\delta-\frac{1}{2\delta}<\delta$$ Then, $$|f(x)-f(y)|=|x^n-y^n|=|\delta^n-\frac{1}{2^n\delta^n}|\geq 1$$ for any $\delta$ as $n\rightarrow \infty$.
Therefore, since no $\delta$ can satisfy the uniform continuity condition for this $\varepsilon$, then we conclude that the function $f(x)=x^n$ is not uniformly continuous.