<s>[INST] Is $f(x) = \sqrt[3]{x}$ continuous on $ [0,\infty)$ and is it uniformly continuous on $[0,\infty)$ Note, there is a similar question here:  Show $\sqrt[3]{x}$ is or isn't uniformly continuous.
I am asking my question anyway because the one here does not ask about the Sequential Criterion for Absence of Uniform continuity
$ f(x)=\sqrt[3]{x}$ is continuous on $[0,\infty)$, note 
$|f(x)-f(c) | \rightarrow |\sqrt[3]{x}-\sqrt[3]{c}| <\epsilon$
$|\sqrt[3]{x}-\sqrt[3]{c}|= |(x-c)(\sqrt[3]{x^2}+\sqrt[3]{xc}+\sqrt[3]{c^2})| $
If we insist on $\delta < 1$, then we ensure that $x$ is within $(c-1,c+1)$
Then $(\sqrt[3]{x^2}+\sqrt[3]{xc}+\sqrt[3]{c^2})| $  $\geq$  $3\sqrt[3]{c^2}$, giving us $|(x-c)(3\sqrt[3]{c^2})|<\epsilon$. 
So $\delta = \dfrac{\epsilon} {3\sqrt[3]{c^2}}$
Is it uniformly continuous on $[0,\infty)$?
Here I use the Sequential Criterion for Absence of Uniform Continuity which states:
A function $f: A \rightarrow \mathbb{R}$ fails to be uniformly continuouson $A$ if and only if $\exists \epsilon_0 >0$ and two sequences $(x_n), (y_n) \in A$ satisfying $|x_n - y_n| \rightarrow 0$ but $|f(x_n) - f(y_n)| \geq \epsilon_0$
I let $x_n=n, y_n=n+\frac{1}{n}$ and observe that $|x_n - y_n| \rightarrow 0$
but
$$|f(x_n) - f(y_n)| = \left| \left( \sqrt[3]{n}-\sqrt[3]{\left(n+\frac{1}{n} \right) }\, \right) \left\{ \frac{\sqrt[3]{n^2} + \sqrt[3]{n\left( n + \frac 1 n \right) + \sqrt[3]{\left( n + \frac 1 n \right)^2}}}{\sqrt[3]{n^2} + \sqrt[3]{ n \left( n + \frac 1 n \right) + \sqrt[3]{\left( n + \frac 1 n \right)^2}}} \right\} \right| $$
$$= \frac{|n-(n + \frac{1}{n})|}{\sqrt[3]{n^2}+\sqrt[3]{n(n+\frac{1}{n})}+\sqrt[3]{(n+\frac{1}{n}})^2},$$
which goes to $0$ and thus $f$ is uniformly continuous on $[0,\infty)$.
 [/INST] Regarding uniform continuity...
You have shown that there is a single pair of sequences for which $|x_n - y_n| \rightarrow 0$ and not $|f(x_n) - f(y_n)| \geq \epsilon_0$.  You have not shown this for any other sequence.  Since you need to show this for any possible pair of sequences, you have only begun your task.
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