Can anyone verify the steps in my proof are correct?
Scratch work: $|f(x)-f(y)| =|x^3-y^3|=|x-y||x^2+xy+y^2|$ (at this point note that we already know $|x-y|<\delta$ for some $\delta>0$) $$|x-y||x^2+xy+y^2|<\delta|x^2+xy+y^2| $$ We now use the triangle inequality several times to find: $$\delta|x^2+xy+y^2| \le \delta(|x^2| +|x||y|+|y^2|) $$ Looking at our domain, we know that at most $|x^2|=|x||y|=|y^2|=1$. So, $$\delta(|x^2| +|x||y|+|y^2|) \le 3\delta $$
Proof: We wish to prove that $\forall \varepsilon>0 \text{ }\exists \delta>0 \text{ } \forall x,y\in[-1,1],|x-y|<\delta:|f(x)-f(y)|<\varepsilon$
$\varepsilon >0$ is given, we choose $\delta = \varepsilon/3$ then $x,y$ satisfying $|x-y|<\delta=\varepsilon/3$ are given.
$$|f(x)-f(y)|=|x^3-y^3|=|(x-y)(x^2+xy+y^2)|=|x-y||x^2+xy+y^2|<\delta|x^2+xy+y^2| $$ By the triangle inequality $$\delta|x^2+xy+y^2| \le \delta(|x^2|+|x||y|+|y^2|)\le3\delta \le3(\varepsilon/3)=\varepsilon $$
Hence $f$ is uniformly continuous on the domain $[-1,1]$