Inequality problem Prove that if 
$$|x-x_0| < \min\left(\frac{\epsilon}{2(|y_0| + 1)}, 1\right)$$ and 
$$|y-y_0| < \frac{\epsilon}{2(|x_0| + 1)},$$ 
then $|xy - x_0y_0| < \epsilon.$
I am doing some problems in Spivak's Calculus on inequalities and came across this problem. Currently I have a sketch of a solution that breaks down the problem into many cases and it is kind of long and messy. I thought maybe someone here could provide a clean and easier solution? If there is a nice solution please tell me a bit behind the thought process (like how you came up with it), instead of giving it as it is.
 A: $$\begin{align*}
|xy-x_0y_0| &= |xy-xy_0 + xy_0 - x_0y_0|\\
&= |x(y-y_0) + (x-x_0)y_0|\\
&\leq |x(y-y_0)| + |y_0||x-x_0|\\
&= |(x-x_0)(y-y_0) + x_0(y-y_0)| + |y_0||x-x_0|\\
&\leq |x-x_0||y-y_0| + |x_0||y-y_0| + |y_0||x-x_0|\\
&\leq |y-y_0| + |x_0||y-y_0| + |y_0||x-x_0| \\
&= (1+|x_0|)|y-y_0| + |y_0||x-x_0|\\
&\lt (1+|x_0|)\left(\frac{\epsilon}{2(|x_0|+1)}\right) + |y_0|\left(\frac{\epsilon}{2(|y_0|+1)}\right)\\
&= \frac{\epsilon}{2} +\left(\frac{|y_0|}{|y_0|+1}\right)\left(\frac{\epsilon}{2}\right)\\
&\lt \frac{\epsilon}{2} + \frac{\epsilon}{2}\\
&= \epsilon.
\end{align*}$$
We can do the step in line $6$, because $|x-x_0|\lt 1$; we can do the step in line $8$ because $|x-x_0|\lt \frac{\epsilon}{2(|y_0|+1)}$ and $|y-y_0|\leq\frac{\epsilon}{2(|x_0|+1)}$. We can do the step in line 10 because $\frac{|y_0|}{|y_0|+1}\lt 1$. 
A: Here is another slightly less familiar trick.
$$xy-x_0y_0=(x-x_0)(y-y_0)+y_0(x-x_0)+x_0(y-y_0)$$
Natural, I think, for this problem, since we have some information about $|x-x_0|$ and $|y-y_0|$ in terms of $y_0$ and $x_0$.
