How can I prove $|x|-|y|\le|x-y|$? You don’t say whether the numbers are real or complex. Assuming that $|x|\ge|y|$ (otherwise the result is trivial) the inequality is equivalent to $|x|\le|y| + |x-y|$.
Let $z = x-y$, then $x = z+y$ and the inequality is equivalent to $|y+z|\le|y| + |z|$. This is known as the triangle inequality and, for complex numbers, it is clear geometrically as the length of each side of a triangle is less than the sum of the other two sides. In the special case of real numbers (or whenever y and z are on the same straight line), the triangle degenerates to a line, but the result is still true.
Note that if $|y|\ge|x|$ then, by the same reasoning, $|y| - |x|\ge|y-x|$. Combining these cases,$ | |x| - |y| |\le|x-y|\le|x| + |y|.$ (The last part follows by putting $w = -y$ in the triangle inequality.)

Given that the triangle inequality holds, 
$$
|x| = |x + y - y| \leq |x+y| + |-y| = |x+y| + |y|
$$
Which is essentially the proof that you argued without the introduction of the new label $z$, and it works on any normed vector space. 
