I had posted a portion of this earlier asking about how to interpret min(). I received some excellent answers, however, I have run into problems and feel stuck. I am posting the question in its entirety.
Let $ x, x_{0},y,y_{0} $ be real numbers and $ \varepsilon $ be a positive real number. If we have
$$|x-x_{0}|<\min \left(\frac{ \varepsilon }{2(|y_{0}|+1)}, 1\right) \text{ and } |y-y_{0}|< \frac{ \varepsilon }{2(|x_{0}|+1)},$$ then prove that $ |xy-x_{0}y_{0}|<\varepsilon $.
A hint is provided:
Write $xy-x_{0}y_{0}$ in terms of $x-x_{0} $ and $y-y_{0} $ and use the triangle inequality twice.
I've been rearranging and writing out what I know etc. in an attempt to find a solution:
$$ |x-x_{0}|(2|y_{0}|)+2|x-x_{0}|<\varepsilon $$
$$ |y-y_{0}|(2|x_{0}|)+2|y-y_{0}|<\varepsilon $$
$$ |x-x_{0}|< 1 $$
Try to rewrite your initial expression as follows:
$$\begin{eqnarray}|xy-x_0y_0|&\leq&|xy-x_0y+x_0y-x_0y_0|\\ &\leq&|y||x-x_0|+|x_0||y-y_0|\\ &\leq&(|y_0|+|y-y_0|)|x-x_0|+|x_0||y-y_0|\\ &\leq& |y_0||x-x_0|+|y-y_0|(|x_0|+|x-x_0|)\\ &<&|y_0|\frac{\varepsilon}{2(|y_0|+1)}+\frac{\varepsilon}{2(|x_0|+1)}(|x_0|+1)\\ &<&2\frac{\varepsilon}{2}=\varepsilon. \end{eqnarray}$$
And between the passages we have used the fact that the condition $$|x-x_0|< \min\left(\frac{\varepsilon}{2(|y_0|+1)},1\right),$$ implies both $$|x-x_0|<\frac{\varepsilon}{2(|y_0|+1)}\quad\text{and}\quad |x-x_0|<1.$$