Usually I would never use the $\epsilon$-$\delta$ criterion for problems of this kind. However, I want to be sure if I have correctly understood the $\epsilon$-$\delta$ criterion.
Let be $f:\mathbb{R}^2 \to \mathbb{R}$, with $f{x \choose y}= x^2+y$. We assume that $\mathbb{R}^2$ is equipped with the maximum norm $\Vert \cdot \Vert_{\infty}$. I want to check if $f$ is continuous at point ${a \choose b}$:
So let be $\epsilon>0$ and ${x \choose y}\in\mathbb{R}^2$ with $\Vert {x \choose y}-{a \choose b}\Vert_{\infty}<\delta$. Let's assume without loss of generality that $\Vert {x \choose y}-{a \choose b}\Vert_{\infty}= |(x-a)|<\delta$. Plugging the points into $f$ yields:
$\vert x^2+y-a^2-b\vert=\vert(x-a)(x+a)+(y-b)\vert\leq \vert(x-a)(x+a)\vert +\vert(y-b)\vert \leq \delta\vert (x+a)\vert +\delta\leq \delta\vert (a+a+\delta)\vert +\delta$ $\leq 2|a|\delta+\delta^2+\delta= \delta^2+\delta(2|a|+1)$.
If I choose the $\delta$ small engough, $0<\delta<1$, then I can further conclude that: $\delta^2+\delta(2|a|+1)< \delta(2|a|+2)$.
Then, I simply set my $\delta$ such that $\delta:= \frac{\epsilon}{(2|a|+2)}$. This means that we have found a $\delta$ which only depends on $a$ and $\epsilon$ and it holds that $\vert f{a \choose b} -f{x \choose y}\vert < \epsilon$ for all points ${x \choose y}$ with $\Vert {x \choose y}-{a \choose b}\Vert_{\infty}<\delta$. So $f$ is conitinuous at point ${a \choose b}$.
Is this correct? Are the upper bounds I used correct? Would you do it differently if you use the $\epsilon$-$\delta$ criterion?
Any comments or feedback is appreciated :)
1) You say: "let's assume without loss of generality that $\|(x,y) -(a,b)\|_\infty = |x-a|$. You can't really do this WLOG (at least I don't see how you don't lose generality), but more importantly you don't need to make this assumption, it isn't used anywhere. By the definition of the $\|\cdot \|_\infty$ norm, $|x-a|, |y-b|<\delta$.
2) You write that you define $\delta := \frac{\epsilon}{2|a| + 2}$, but you really need to take $\delta := \min( \frac{\epsilon}{2|a| + 2}, 1)$ since you used, $\delta \leq 1$ above.
Other than that, everything looks good.