The statement, and your proof of it, are correct.
There is also a stronger form of this fact: if a function is uniformly continuous on a dense set, then it has a (unique) continuous extension to the entire space, which is also 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
You can prove it by contradiction. Suppose you have a universal $\delta$ for some fixed $\epsilon>0$. From your calculations you know that $$ |g(t)-g(a)|=|t-a|/at $$ If you choose $a<\min\{\delta,1/2\epsilon\}$ and $t=a/2$ (thus $|t-a|<\delta$), you get $$ |g(t)-g(a)|=1/a>2\epsilon, $$ a contradiction