$P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous. I need to prove $P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous.
I just proved that the sum is continuous, but I'm lost at how to manipulate the inequalities in this case.
My attempt:
Let $\epsilon>0$.
$d((x,y),(a,b)) = \sqrt{(x-a)^2 + (y-b)^2} $. So I need to find some $\delta$ such that $\sqrt {(x-a)^2 + (y-b)^2} < \delta \Rightarrow |xy-ab| < \epsilon $.
I verified that $\sqrt{(x-a)^2 + (y-b)^2} \geq |x-a| $ and $\sqrt{(x-a)^2 + (y-b)^2} \geq |y-b|$, but can't see if that helps. I don't know how can I find $\delta$ in this case, can someone give me a hint?
Thanks.
 A: Fix $a,b\in\mathbb{R}$, then we have the following:
Scratch work: If $d((x,y),(a,b))=|x-a|+|y-b|<\delta$ then $|x-a|<\delta$ and $|y-b|<\delta$. So we have:
\begin{align*}
x-a&<\delta\\
x+a&<\delta+2a\\
\end{align*}
and so we have:
\begin{align*}
|(x+a)(y-b)|&=|xy-ab + (ay-bx)|\geq |xy-ab|\\
&\Downarrow\\
|xy-ab|&\leq |(x+a)(y-b)|\\
&|xy-ab|<(\delta+2a)\cdot \delta :=\epsilon
\end{align*}
So, the appropriate $\delta$ might be the solution of the equation
\begin{align*}
\delta^2+2a\delta&=\epsilon\\
(\delta+a)^2&=\epsilon+a^2\\
\delta&=\sqrt{\epsilon+a^2}-a
\end{align*}
Proof of continuity: Take $\epsilon>0$ arbitrary, and put $\delta=\sqrt{\epsilon+a^2}-a$. Then if $d((x,y),(a,b))=|x-a|+|y-b|<\delta$ we have $|y-b|<\delta$ and $|x+a|<\delta+2a$, which implies:
\begin{align*}
|xy-ab|&\leq |xy-ab+ay-bx| &\text{(by triangle inequality)}\\
&=|(x+a)(y-b)|<(\delta+2a)\cdot \delta &\\
&=\left(\left(\sqrt{\epsilon+a^2}-a\right)+2a\right)\cdot \left(\sqrt{\epsilon+a^2}-a\right) &\text{(replacing $\delta=\sqrt{\epsilon+a^2}-a$)}\\
&=\left(\sqrt{\epsilon+a^2}+a\right)\cdot \left(\sqrt{\epsilon+a^2}-a\right)\\
&=\left(\sqrt{\epsilon+a^2}\right)^2 - a^2\\
&=\epsilon+a^2-a^2=\epsilon.
\end{align*}
So, since for every $\epsilon>0$ there is $\delta>0$ such that $d((x,y),(a,b))<\delta$ implies $|xy-ab|<\epsilon$, we conclude that the function $P(x,y)=x\cdot y$ is continuous in $(a,b)$. 
Answer after edited question:
Notice that $d((x,y),(a,b))=\sqrt{(x-a)^2+(y-b)^2}<\delta$ implies
\begin{align*}
(x-a)^2<\delta^2 &\Rightarrow |x-a|<\delta\\
(y-b)^2<\delta^2 &\Rightarrow |y-b|<\delta
\end{align*}
So the previous proof still applies for the edited question.
A: Hint:
$(x+h)(y+g) - xy =  xg + yh + hg$
For example with $\eta = \frac{min(\epsilon,1)}{3max(|x|,|y|,1)}$
you can show that
$d((x+h,y+g),(x,y)) \lt \eta \Rightarrow |(x+h)(y+g) - xy| \lt  \epsilon$
A: Since
$$
\underbrace{(x_1^2+x_2^2)}_{\|x\|^2}\underbrace{(y_1^2+y_2^2)}_{\|y\|^2}=\underbrace{(x_1y_1+x_2y_2)^2}_{(x\cdot y)^2}+(x_1y_2-x_2y_1)^2
$$
We have
$$
|x\cdot y|\le\|x\|\,\|y\|
$$
Thus,
$$
\begin{align}
\left|\,x\cdot y-a\cdot b\,\right|
&=\left|\,(x-a)\cdot b+x\cdot(y-b)\,\right|\\
&\le\|x-a\|\,\|b\|+\|y-b\|\,\|x\|
\end{align}
$$
If we have $\|x-a\|\le\min\left(\frac{\epsilon}{2\|b\|},\frac{\|a\|}2\right)$ and $\|y-b\|\le\frac{\epsilon}{3\|a\|}\le\frac{\epsilon}{2\|x\|}$, then $\left|\,x\cdot y-a\cdot b\,\right|\le\epsilon$.
Therefore, if we let $\delta=\min\left(\frac{\epsilon}{2\|b\|},\frac{\|a\|}2,\frac{\epsilon}{3\|a\|}\right)$, and $\sqrt{\|x-a\|^2+\|y-b\|^2}\le\delta$, then $\left|\,x\cdot y-a\cdot b\,\right|\le\epsilon$.
