I don't think so. Every function which is differentiable and has bounded derivative is uniformly continuous (Wikipedia), and it is possible to have a function that is above y = x and has a bounded derivative. The problem with $\displaystyle x^3$ is that its derivative is not bounded on all $\displaystyle \mathbb{R}$. In fact, the link above has an outline of the proof that $\displaystyle x^2$ is not uniformly continuous.

This is correct, but you are interested in the lower, not upper bound on $\displaystyle |x^3-y^3|$: you want to show that it is $\displaystyle \ge\epsilon$.so we must prove that there is $\displaystyle \epsilon>0$ so for all $\displaystyle \delta>0$

there is x,y in R which $\displaystyle |x-y|<\delta$ and$\displaystyle |f(x)-f(y)|=>\epsilon$

$\displaystyle |x^3-y^3|=|(x-y)(x^2+xy+y^2)|<|\delta (x^2+xy+y^2)|$