If $x,y$ $\in$ $\mathbb{R}$, then $||x|-|y|| \le |x|-|y|$ If $x,y$ $\in$ $\mathbb{R}$, then $||x|-|y|| \le |x|-|y|$
I'm supposed to justify the previous statement by citing a theorem, giving a proof, or giving a counterexample. 
The following is a proof of the statement.
$|x| = ||x|| = ||x|-|y|+|y|| = ||x|-|y||+||y|| \le |x|$.
Therefore, $||x|-|y|| \le |x| - |y|$
I would like to know if this is sufficient, and how this proof can be improved.
 A: As pointed out by Dr. Sonnhard Graubner, the statement
$\vert \vert x \vert - \vert y \vert \vert \le \vert x \vert - \vert y \vert \tag 1$
is false.  A similar looking statement,
$\vert \vert x \vert - \vert y \vert \vert \le \vert x - y \vert, \tag 2$
is however true, and the proof looks similar to what was given in the text of the question:
$x = y + (x - y); \tag 3$
$\vert x \vert = \vert y + (x - y) \vert \le \vert y \vert + \vert x - y \vert, \tag 4$
by the triangle inequality ($\vert x + y \vert \le \vert x \vert + \vert y \vert$); thus
$\vert x \vert - \vert y \vert \le \vert x - y \vert; \tag 5$
interchanging the roles of $x$ and $y$ we see that
$\vert y \vert - \vert x \vert \le \vert y - x \vert; \tag 6$
noting that $\vert x - y \vert = \vert y - x \vert$ we thus, from (5) and (6) obtain
$-\vert x - y \vert = -\vert y - x \vert \le \vert x \vert - \vert y \vert \le \vert x - y \vert, \tag 7$
which is equivalent to
$\vert \vert x \vert - \vert y \vert \vert \le \vert x - y \vert. \tag 8$
This result is sometimes known as the reverse triangle inequality.
Perhaps our OP was reaching for something along these lines.
A: Just substitute $$x=1,y=4$$ then we get
$$|1-4|\le 1-4$$?
A: The true inequality (valid for all $x,y$) is
$$\bigl||x|-|y|\bigr|\le|x-y|.$$
On the contrary, 
we have $$\bigl||x|-|y|\bigr|\ge |x|-|y|.$$
