Critique my proof of $\lvert x \rvert + \lvert y \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert, \ x,y \in \mathbb{R}$ Critique my proof of $\lvert x \rvert + \lvert y \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert, \ x,y \in \mathbb{R}$
Let $x, y \in \mathbb{R}$.
Case #1: $x = y$
$\Rightarrow \lvert x \rvert = \lvert y \rvert$. Hence,
$$\lvert x + y \rvert = \lvert x \rvert + \lvert x \rvert = 2 \lvert x \rvert$$
and $$\lvert x - y \rvert = \lvert x - x\rvert = 0.$$
Thus, $\lvert x \rvert$ + $\lvert y \rvert = 2 \lvert x \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert = 2 \lvert x \rvert$.
Case #2: $x > y$
Assume for the sake of contradiction that $\lvert x \rvert + \lvert y \rvert > \lvert x + y \rvert + \lvert x - y \rvert.$
Hence, $$ \lvert x + y \rvert + \lvert x - y \rvert < \lvert x \rvert + \lvert y \rvert$$
$$\leq \lvert x \rvert + \lvert y \rvert + \lvert x - y \rvert < \lvert x \rvert + \lvert y \rvert$$
by the triangle inequality, which is a contradiction b/c $\lvert x - y \rvert > 0$. Thus, it must be that $\lvert x \rvert + \lvert y \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert$ in this case.
Case #3: $x < y$
WLOG, the proof is the same as case #2.
$\therefore \lvert x \rvert + \lvert y \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert, \ x,y \in \mathbb{R}$.
 A: In the second case, there is a mistake because from the inequalities
$$\begin{cases}|x+y|+|x-y|<|x|+|y|\\|x+y|\leqslant|x|+|y|\end{cases}$$
you cannot deduce that
$$|x|+|y|+|x-y|<|x|+|y|\;.$$
Indeed, $\,A+B<C+D\,$ and $\,A\leqslant C+D\,$ do not imply that $\,C+D+B<C+D\,.$
For example, if $\,A=1,B=2,C=3,D=4\,,\,$ then $\,A+B<C+D\,$ and $\,A<C+D\;\;$ but $\,C+D+B\not\lt C+D\,.$

Addendum 1:
You could prove the property in the following way.
If $\,x\geqslant0\,$ and $\,y\geqslant0\,,\,$ then
$|x|+|y|=x+y=|x+y|\leqslant|x+y|+|x-y|\,.$
If $\,x<0\,$ and $\,y<0\,,\,$ then
$|x|+|y|=-x-y=-(x\!+\!y)=|x+y|\leqslant|x+y|+|x-y|\,.$
If $\,x\geqslant0\,$ and $\,y<0\,,\,$ then
$|x|+|y|=x-y=|x-y|\leqslant|x+y|+|x-y|\,.$
If $\,x<0\,$ and $\,y\geqslant0\,,\,$ then
$|x|+|y|=-x+y=-(x\!-\!y)=|x-y|\leqslant|x+y|+|x-y|\,.$

Addendum 2:
A better way to prove the property is the following one.
$\begin{align}|x|+|y|&=\left|\dfrac{(x+y)+(x-y)}2\right|+\left|\dfrac{(x+y)-(x-y)}2\right|\leqslant\\&\leqslant\dfrac{|x+y|+|x-y|}2+\dfrac{|x+y|+|x-y|}2=\\&=|x+y|+|x-y|\;.\end{align}$
A: There is a flaw in your case 2 because you used the triangle inequality the wrong way. $|x + y|+|x - y|<blah$ and $|x + y|\le|x|+|y|$ don't imply
$|x|+|y|+|x - y|<blah.$
