Sequences and inequalities in $\mathbb{R}^n$ Let $x_k,y_k$ be sequences in $\mathbb{R}^n$ with $\lim x_k=a$, $\lim y_k=b$ and $$|x_k-b|<r<|y_k-a|$$ prove that $|a-b|=r.$
My attempt was:
We can write $|x_k-b+(a-a)|=|x_k-a+(a-b)|$ and $|y_k-a+(b-b)|=|y_k-b +(b-a)|$. My first question is: In $\mathbb{R}$ we can write $|u|-|v|\leq |u-v|$, can we do $$|x_k-a|-|b-a|\leq |x_k-a-(b-a)| ?$$
if we can, we have $$|x_k-a|-|b-a|<r<|y_k-b|+|b-a|$$
and $|x_k-a|<\epsilon,|y_k-b|<\epsilon.$ So we can write $$\epsilon-|b-a|<r<\epsilon +|b-a|$$ This was my last "step". I can't see how can i conclude that $|b-a|=r.$

Your second last and last inequalities are true but not useful. Instead you want to use triangle inequality at the second last inequality as follows:
$|b-a|-\epsilon < |b-a|-|x_k-a|<r < |y_k-b|+|b-a|< |b-a|+\epsilon\implies |r-|b-a||<\epsilon\implies r = |b-a|$ since it is true for any $\epsilon > 0$.
