Prove that $\left|(|x|-|y|)\right|\leq|x-y|$ 
Prove that $\left|(|x|-|y|)\right|\leq|x-y|$


Proof:
$$\begin{align} \left|(|x|-|y|)\right| &\leq|x-y| \\
{\left|\sqrt{x^2}-\sqrt{y^2}\right|}&\leq \sqrt{(x-y)^2}   &\text{($\sqrt{a^2}=|a|)$}\\
\sqrt{\left(\sqrt{x^2}-\sqrt{y^2}\right)^2}&\leq \sqrt{(x-y)^2} &\text{($\sqrt{a^2}=|a|)$} \\
\left(\sqrt{x^2}-\sqrt{y^2}\right)^2  &\leq (x-y)^2&\text{$(0\leq a\leq b\implies a^2\leq b^2)$}\\
\left(\sqrt{x^2}\right)^2-2\sqrt{x^2}\sqrt{y^2}+\left(\sqrt{y^2}\right)^2&\leq x^2-2xy+y^2 &\text{(Distribution)} \\
|x|^2-2|x||y|+|y|^2&\leq x^2-2xy+y^2 &\text{($\sqrt{a^2}=|a|)$}\end{align} \\ $$
From here, we can see that $|x|^2+|y|^2=x^2+y^2$, and it suffices to show that the following holds: $$-2|x||y|\leq -2xy.$$
When $x>0,y>0$ or $x<0,y<0$, the product $xy$ is positive, and so 
$-2|x||y|=-2xy$ holds.
When $x<0,y>0$ or $x>0,y<0$, the product $xy$ is negative, and so $-2xy>0$, and it follows that:
$$-2|x||y|<0<-2xy. $$
$\square$
Is this proof valid? I am unsure if I overlooked anything, also I have some specific questions about the proof (if it is valid up to this point). From the 3rd to the 4th inequality, it feels like I am making a leap in knowing that $a\leq b$ prior to proving what's to be proved by using the result $a^2\leq b^2$. Is this a problem?
What are some alternative ways of showing this inequality?

Because of the triangle inequality, there is: $\left| x+y \right| \le \left| x \right| +\left| y \right| $. Using this fact:
\begin{align}
\left| x \right| &= \left| (x-y)+y \right| \\
&\le \left| x-y \right| +\left| y \right|   \\
\left| x \right| -\left| y \right| &\le \left| x-y \right| \tag{1}
\end{align}
Proceeding similarly: 
\begin{align}
\left| y \right| &= \left| (y-x)+x \right| \\
\left| y \right| &\le \left| y-x \right| +\left| x \right| \\
\left| y \right| -\left| x \right| &\le \left| y-x \right|  \tag{2} \\ 
\end{align}
And finally combining $(1)$ and $(2)$ :$$\left|(|x|-|y|)\right|\leq|x-y|$$
