There is a proof that shows that $f(x)=x^n$ is not uniformly continuous on $[0,\infty[$ in the following way:
Choose $\epsilon=1$. Set $y=x+\delta/2$. It follows that $|x-y|<\delta$ for all $x,y\in[0,\infty[$.
Now $|x^n+y^n|=|x^n-(x+\delta/2)^n|=(x+\delta/2)^n-x^n\ge(x+\delta/2)x^{n-1}-x=\frac{\delta}{2}x^{n-1}$.
Since I can choose any $x$ let's take $x=2/\delta^{n-1}$ and we have proven that $|x^n-y^n|\ge 1=\epsilon$ and therefore is not uniformly continuous.
I know now that if I choose an any interval, lets say $[0,1]$, then $f$ is uniformly continuous. But I was asking myself if I would believe that $f$ is not uniformly cont. on that interval, why would the proof above would not work?
At the very ending, when I choose my $x$ value, I take not account all values of $\delta$, that since $|x|\le1$, I restrict $\delta$ to $\delta\ge2^{1/(n-1)}$ and therefore me trying to disproof uniform continuity on the interval is not sufficient since I have not proven it for ALL $\delta$.
Is this conclusion right, or am I completely lost, and if so, why wouldn't this proof work?