Normed vector space inequality $|\|x\|^2 - \|y\|^2| \le \|x-y\|\|x+y\|$

Let $x=u+v$ and $y=u-v$, then $$|\|x\|^2-\|y\|^2|=|\|u+v\|^2-\|u-v\|^2|=4|u^\top v|$$ Then replace $u=\frac{x+y}{2}$ and $v=\frac{x-y}{2}$ in the above equation and you obtain $$4|u^\top v|=|(x+y)^\top (x-y)|\leq \|x+y\|\|x-y\|$$ and the proof is complete.

We may assume w.l.o.g. that $\|x\|^2 \geq \|y\|^2$. Write $x = u + v$ and $y = u - v$. Now the inequality can be rewritten as $$ \|u + v\|^2 \leq 4 \|u\| \|v\| + \|u - v\|^2. $$ But this is the inequality one gets by combining $\|u + v\|^2 \leq (\|u\| + \|v\|)^2$ and $|\|u\| - \|v\||^2 \leq \|u - v\|^2$.