I would like to ask whether the following proof is correct and, in particular, why the step (*) of this proof is valid?
The problem is to show that the function $+: \mathbb{R}^2 \to \mathbb{R}, (x,y)\mapsto x+y$ is uniformly continuous and give a counterexample for $\times: \mathbb{R}^2 \to \mathbb{R}, (x,y)\mapsto xy$.
addition $f(x,y)=x+y$
Given $\varepsilon>0,$ let $\delta=\varepsilon/2.$ Then for any $(x,y)$, $(u,v)$,
$d((x,y),(u,v))<\delta$
$\implies|x-u|<\varepsilon/2\,,\,|y-v|<\varepsilon/2$ (*)
$\hspace{0ex}\implies d(f(x,y),f(u,v))=|x+y-(u+v)|$
$\hspace{8ex}=|x-u+y-v|$
$\hspace{8ex}\leq|x-u|+|y-v|<\varepsilon/2+\varepsilon/2=\varepsilon$
multiplication $g(x,y)=xy$
Given $(x,y)\,, x>0,y>0,$ and $\delta>0,$ observe that
$\hspace{0ex}|g(x,y)-g(x+\delta,y+\delta)|=|xy-(x+\delta)(y+\delta)|$
$\hspace{22ex}=(x+y)\delta+\delta^2$
$\hspace{22ex}>(x+y)\delta\,.$
Consequently, given $\varepsilon>0\,,\,\,$ for any $\delta>0$ , if we choose x and y positive with $x+y>\varepsilon/\delta\,\,,\, $then $|xy-(x+\delta)(y+\delta)|>\varepsilon.$