Need help with a proof about absolute value 
Prove $||x| - |y|| \le |x-y|$

Here's my attempt of the proof:
Since $x-y = |x-y|$ or $x-y = -|x-y|$, then $-|x-y| \le x-y \le |x-y|$. 
Also, $|x| = |-x|$ and $|y| =|-y|$, so we have that $|x|-|y| \le |x-y|$. 
We know that for any $\epsilon \gt 0$, $|z| \le \epsilon$ iff $-\epsilon \le z \le \epsilon$. 
Let $z = |x|-|y|$ and take $\epsilon = |x-y|$.
This implies $||x| - |y|| \le |x-y|$.
I want to know if I missed any key details that I should've written in my proof, and above all, is my proof correct?
 A: Ah yes, the reverse triangle inequality, appealing to the intuition insofar as it confirms that the absolute difference in the magnitudes $\vert x \vert$, $\vert y \vert$ of $x$ and $y$ is bounded by their difference $\vert x - y \vert$ itself:  if $x$ is close to $y$, then the size of $x$ is close to the size of $y$:
$\vert \vert x \vert - \vert y \vert \vert \le \vert x - y \vert; \tag 0$
since our OP Ash's work on proving this has been adequately debriefed in the comments to the question itself, I simply present my own demonstration, merely a re-iteration of a very standard approach: 
$\vert y \vert = \vert x + y - x \vert \le \vert x \vert + \vert y - x \vert = \vert x \vert + \vert x - y \vert; \tag 1$
so
$\vert y \vert - \vert x \vert \le \vert x - y \vert; \tag 2$
we may reverse the roles of $x$ and $y$ in the above we obtain
$\vert x \vert - \vert y \vert \le \vert x - y \vert; \tag 3$
we may multiply (2) by $-1$ and find
$-\vert x - y \vert \le \vert x \vert - \vert y \vert; \tag 4$
we combine (3) and (4):
$-\vert x - y \vert \le \vert x \vert - \vert y \vert \le \vert x - y \vert, \tag 5$
which by definition is equivalent to
$\vert \vert x \vert - \vert y \vert \vert \le \vert \vert x - y \vert \vert = \vert x - y \vert. \tag 6$
$OE\Delta$.
A: Your question
$$\vert\vert x\vert-\vert y\vert\vert\le\vert x-y\vert$$
Square both sides
$$(\vert x\vert-\vert y\vert)^2\le(x-y)^2$$
Distribute
$$\vert x\vert^2 + \vert y\vert^2 - 2\vert x\vert\vert y\vert\le x^2+y^2-2xy$$
Simplify
$$x^2 + y^2 -2\vert x\vert\vert y\vert\le x^2+y^2-2xy$$
$x^2+y^2$ Cancel out
$$-2\vert x\vert\vert y\vert\le-2xy$$
Divide by $-2$
$$\vert x\vert \vert y\vert\ge xy$$
Which is the same as
$$\vert xy\vert\ge xy$$
And to finish this is
$$\vert a\vert\ge a$$
Thanks, I had a lot of fun answering this
