Prove that if $|x-x_0|<\min\left(\dfrac{\varepsilon}{2\left(|y_0|+1\right)},1\right)\quad$ and $\quad|y-y_0|<\dfrac{\varepsilon}{2\left(|x_0|+1\right)}$, then $|xy -x_0y_0|<\varepsilon$.
\begin{align} |y-y_0| &< \frac{\varepsilon}{2(|x_0|+1)}\\ |y-y_0|(|x_0|+1)&< \varepsilon/2\\ |y-y_0|(|x_0|+|x-x_0|)&< \varepsilon/2\\ |y-y_0||x_0+(x-x_0)|&<\varepsilon/2\\ |y-y_0||x|&<\varepsilon/2\\ |xy-xy_0|&<\varepsilon/2\\ |xy-xy_0|-|x-x_0|&<\varepsilon/2 \end{align}
and
\begin{align} |x-x_0|&<\frac{\varepsilon}{2(|y_0|+1)}\\ (|y_0)|+1)|x-x_0|&<\varepsilon/2\\ |y_0||x-x_0|+|x-x_0|&<\varepsilon/2\\ |y_0x-y_0x_0|+|x-x_0|&<\varepsilon/2\\ \end{align}
Adding the first and second results: \begin{align} |xy-xy_0|-|x-x_0|+|y_0x-y_0x_0|+|x-x_0|&<\varepsilon\\ |xy-xy_0|+|y_0x-y_0x_0|<\varepsilon\\ |(xy-xy_0)+(y_0x-x_0y_0)|<\varepsilon \\|xy-x_0y_0|<\varepsilon \end{align}
$$|x-x_0|<\min \left(\frac{\varepsilon}{2(|y_0|+1)},1\right)$$ just means that $|x-x_0|<\frac{\varepsilon}{2(|y_0|+1)}$ and $|x-x_0|<1$ at the same time. Now, since $$|x|-|x_0|\le |x-x_0|<1$$ we can say that $|x|<1+|x_0|$. Finally, note that $$\begin{align} |xy-x_0y_0|&=|x(y-y_0)+y_0(x-x_0)| \\ &\leq |x||y-y_0|+|y_0||x-x_0| \\ &<(1+|x_0|)\cdot \frac{\varepsilon}{2(|x_0|+1)}+|y_0|\cdot \frac{\varepsilon}{2(|y_0|+1)} \\ &= \frac{\varepsilon}{2}+\left(1-\frac{1}{|y_0|+1}\right)\cdot \frac{\varepsilon}{2} \\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\ &=\varepsilon. \end{align}$$