Showing $f(x)=x^4$ is not uniformly continuous

To show that it is not uniformly continuous on the whole line, there are two usual (and similar) ways to do it:

1. Show that for every $\delta > 0$ there exist $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|$ is greater than some positive constant (usually this is even arbitrarily large).
2. Fix the $\varepsilon$ and show that for $|f(x)-f(y)|<\varepsilon$ we need $\delta = 0$.

First way:

Fix $\delta > 0$, set $y = x+\delta$ and check $$\lim_{x\to\infty}|x^4 - (x+\delta)^4| = \lim_{x\to\infty} 4x^3\delta + o(x^3) = +\infty.$$

Second way:

Fix $\epsilon > 0$, thus $$|x^4-y^4| < \epsilon $$ $$|(x-y)(x+y)(x^2+y^2)| < \epsilon $$ $$|x-y|\cdot|x+y|\cdot|x^2+y^2| < \epsilon $$ $$|x-y| < \frac{\epsilon}{|x+y|\cdot|x^2+y^2|} $$

but this describes a necessary condition, so $\delta$ has to be at least as small as the right side, i.e.

$$|x-y| < \delta \leq \frac{\epsilon}{|x+y|\cdot|x^2+y^2|} $$

so if either of $x$ or $y$ tends to infinity then $\delta$ tends to $0$.

Hope that helps ;-)

Edit: after explanation and calculation fixes, I don't disagree with your proof.