Prove that $||x|-|y||\le |x-y|$ I've seen the full proof of the Triangle Inequality 
\begin{equation*}
|x+y|\le|x|+|y|. 
\end{equation*}
However, I haven't seen the proof of the reverse triangle inequality: 
\begin{equation*}
||x|-|y||\le|x-y|.
\end{equation*}
Would you please prove this using only the Triangle Inequality above?
Thank you very much.

Given that we are discussing the reals, $\mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,z\in\mathbb{R}, \quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.
Start with $x=x+0=x+(-y+y)=(x-y)+y$. 
Then apply $|x| = |(x-y)+y|\leq |x-y|+|y|$.   By so-called "first triangle inequality."
Rewriting $|x|-|y| \leq |x-y|$ and $||x|-y|| \leq |x-y|$.
The item of Analysis that I find the most conceptually daunting at times is the notion of order $(\leq,\geq,<,>)$, and how certain sentences can be augmented into simpler forms.
Hope this helps and please give me feedback, so I can improve my skills.
Cheers. 
