Multivariable $\epsilon-\delta$ proof verification Prove that $\lim\limits_{(x,y) \to (1,1)} xy=1$
Of course, I am aware that this is "obvious", but I want to add some rigor to it. When I searched around for multivariable limits using $\epsilon-\delta$, most of the examples had $(x,y) \rightarrow (0,0)$, but in this case I have $x$ and $y$ approaching something else.
$(x,y) \rightarrow (1,1) \Leftrightarrow \lvert\lvert (x,y)-(1,1)\lvert\lvert \rightarrow 0$ which can be written as $0 < \sqrt {(x-1)^2+(y-1)^2} < \delta$ for some arbitrarily small $\delta >0$.
Goal: show that $\forall$ $\epsilon>0$ $\exists$ $\delta>0$ such that 
$0 < \sqrt {(x-1)^2+(y-1)^2}<\delta\Rightarrow0<|xy-1|<\epsilon$
Proof:
If $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$, then
$|xy-1|=|xy-x-y+1+x+y-2|=|(x-1)(y-1)+(x-1)+(y-1)|$
$\le|(x-1)(y-1)|+|x-1| +|y-1|=|x-1||y-1|+|x-1|+|y-1|$
$=(\sqrt{(x-1)^2})(\sqrt{(y-1)^2})+\sqrt{(x-1)^2}+\sqrt{(y-1)^2}$
$\le(\sqrt{(x-1)^2+(y-1)^2})^2+2\sqrt{(x-1)^2+(y-1)^2}<\delta^2+2\delta$
If $\delta=$$\epsilon \over4$, then $\delta^2+2\delta=\frac{\epsilon^2}{16}+\frac{8\epsilon}{16}=\frac{\epsilon^2+8\epsilon}{16}$
Now, $\frac{\epsilon^2+8\epsilon}{16}<\epsilon \Leftrightarrow \epsilon^2+8\epsilon<16\epsilon\Leftrightarrow\epsilon(\epsilon-8)<0$
Since $\epsilon>0$, we have $\epsilon(\epsilon-8)<0$ if and only if $\epsilon<8$
So if we have $\epsilon<8$, then we can pick $\delta=\frac{\epsilon}{4}$ which gives us $\delta^2+2\delta<\epsilon$. If $\epsilon \ge8$, then we can pick $\delta=2$ which gives us $\delta^2+2\delta\le\epsilon$.
Therefore, if we pick $\delta =$ min {${\frac{\epsilon}{4}, 2}$}, then $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$ implies $0<|xy-1|<\epsilon$ 
Thus, $\lim\limits_{(x,y) \to (1,1)} xy=1$

Your proof is correct and nice.
Take $\delta<1$
By doing this a priori you will find the delta you need  more easily and save some time from the calculations
You proved that $$|xy-1| \leq |x-1||y-1|+|x-1|+|y-1|$$
Thus $$|x-1|=\sqrt{(x-1)^2} \leq \sqrt{(x-1)^2+(y-1)^2} < \delta$$
$$|y-1|=\sqrt{(y-1)^2} \leq \sqrt{(x-1)^2+(y-1)^2} < \delta$$
Thus $|f(x,y)-1|=|xy-1| <\delta^2 +2\delta=\delta(\delta+2) <3 \delta$
Choose $0<\delta <\min\{ \frac{\epsilon}{3},1\}$ and you have the result
