If $|x-x_0|

This is a problem from Spivak's Calculus 4th ed., Chapter 1

If $|x-x_0|<\frac {\varepsilon}{2}$ and $|y-y_0|<\frac {\varepsilon}{2}$ then $|(x+y)-(x_0+y_0)|<\varepsilon$ and $|(x-y)-(x_0-y_0)|<\varepsilon$

Here is my proof:

If $|x-x_0|<\frac {\varepsilon}{2}$ and $|y-y_0|<\frac {\varepsilon}{2}$ then $|x-x_0|+|y-y_0|<\varepsilon$ (I've proved that if $a<c$ and $b<d$ then $a+b<c+d$ in a previous exercise so I used that to come to this conclusion).

$|x-x_0|+|y-y_0|\geq|x-x_0+y-y_0|=|(x+y)-(x_0+y_0)|$ (I've proved in a previous exercise that $|x|+|y|\geq|x+y|$ in a previous exercise so I used that to come to this conclusion.

$\therefore$ Since $|(x+y)-(x_0+y_0)|\leq|(x+y)-(x_0+y_0)|<\varepsilon$ it follows that $|(x+y)-(x_0+y_0)|<\varepsilon$.

Is the proof correct?