$x^{1+\epsilon}$ is not uniformly continuous on $[0,\infty)$ There are two questions.
First: is the proof underneath correct?
Let $\epsilon>0$ and let $f(x)=x^{1+\epsilon}$. I aim to show that $f$ is not uniformly continuous on $[0,\infty)$.
We will show that for any $\delta>0$ there exists $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\geq 1$.
Now let $\delta>0$ be given. Consider $f'(x)=(1+\epsilon)x^\epsilon$. We have that $f'\rightarrow\infty$ as $x\rightarrow \infty$ and that $f'$ is monotone increasing. Now choose an $x_0$ large enough so that $f'(x)$ is larger than $2/\delta$ for all $x\geq x_0-\delta$.
Now consider $|f(x_0)-f(y)|$ for some $y\in(x_0+\delta/2,x_0+\delta)$. The Mean Value Theorem guarantees us that there is a $\zeta\in(x_0, y)$ such that $|f(x_0)-f(y)|=|f'(\zeta)||x_0-y|$.
We thus have that
$$
\begin{array}{rl}
|f'(\zeta)||x_0-y|&=(1+\epsilon)\zeta^\epsilon|x_0-y|\\
&\geq(1+\epsilon)\zeta^\epsilon\delta/2\\
&\geq(1+\epsilon)x_0^\epsilon\delta/2\\
&\geq2/\delta\cdot\delta/2\\
&=1
\end{array}
$$
Thus for this choice of $x_0$ and $y$, we have that $|f(x_0)-f(y)|\geq 1$ and $f$ is not uniformly continuous.
Second: If all is well and good, what other ways are there to prove this claim? I feel like the approach I took was a little bulky.

An alternative solution is to find a pair of cofinal sequences $(x_n)$ and $(y_n)$ such that $(x_n^{1+\varepsilon})$ and $(y_n^{1+\varepsilon})$ are not cofinal. Can you do this?
Here cofinal means $d(x_n,y_n)\to 0$. For example, if $\varepsilon=1$; I can take $x_n=\sqrt n$ and $y_n=\sqrt{n+1}$. Then $|x_n-y_n|\to 0$ but $|x_n^2-y_n^2|=1\not\to 0$.
