First of all, you make a mistake many beginners make: when talking about uniform continuity, it is essential to tell on what set you are studying it. For instance, $x^2$ is uniformly continuous on $[0,r]$ for every $r>0$, but is not so on $[0,\infty)$. Similarly, $\sin \frac 1 x$ is uniformly continuous on $[r,\infty)$ for every $r>0$, but is not so on $(0,\infty)$. The domain of definition is crucial.
Second, there are two main alternative techniques to plainly using the definition:
the first one is quick: if $f$ is defined on a compact set and is continuous, then it is uniformly continuous
the second one is subtler: if $f$ is derivable with $f'$ continuous and you manage to find a $M>0$ such that $\sup |f'(x)| \le M$ on your domain of definition, then $f$ will be Lipschitz and, in particular, uniformly continuous.
Note that in the definition of uniform continuity, given $\varepsilon > 0$, you need to provide a $\delta > 0$ that works for all $x,y \in [0,1]$. In particular, you cannot have a $\delta$ that depends on $x$. This is in contrast to proving that $f$ is continuous at a specific $x$ where the $\delta$ you provide can depend on $x$.
In your case, we have
$$ |x^3 - y^3| = |(x - y)(x^2 + xy + y^2)| \leq |x - y||x^2 + xy + y^2|. $$
Now, if $x,y \in [0,1]$, we have $|x^2 + xy + y^2| \leq 3$ so we can deduce that
$$ |x^3 - y^3| \leq 3|x - y|. $$
Hence, given $\varepsilon > 0$, we can take $\delta = \frac{\varepsilon}{3}$ and then if $|x - y| < \delta$ then $|x^3 - y^3| \leq 3|x - y| < 3\delta = \varepsilon$.
Note that the same argument would work if you wanted to prove uniform continuity on $[0,L]$ with the constant $3$ replaced by a different constant $C_L$ which bounds $|x^2 + xy + y^2|$ on $[0,L]$ (for example, $C_L$ can be $3L^2$).
Expressed in this way, we see that your basic intuition is correct. As $L$ gets larger, our constant $C_L$ also gets larger and hence our $\delta = \frac{\varepsilon}{C_L}$ gets smaller and smaller. In the limit, this should lead us to suspect (but this is not a formal proof!) that uniform continuity will fail.
Best Answer
The comments and other answer address the uniform continuity on compact sets. @hermes points out the crux of your proof, that $|x + x_0| < 2$, cannot be extended to all of $\mathbb{R}$. Turns out there is no fix for that.
Let $\varepsilon = 1$. We need to show that for all $\delta > 0$, there exist $x,y$ such that $|x - y| < \delta$ but $|f(x) - f(y)| > 1$. The key is the dependence of $x$ and $y$ on $\delta$. To this end, let $\delta > 0$ be given, and choose $y = x + \frac{\delta}{2}$. Then $|x-y| = \frac{\delta}{2} < \delta$, but $$|f(x) - f(y)| = |x^2 - y^2| = \left|x^2 - \left(x^2 + x\delta + \frac{\delta^2}{4}\right)\right| = \left|x\delta + \frac{\delta^2}{4}\right|,$$ and for sufficiently large (or small) $x \in \mathbb{R}$, we can make the last quantity larger than 1. Hence we do not have uniform continuity on $\mathbb{R}$.