Let $x_n, y_n \in \mathbb{R}$, $\lim x_n=a$, $\lim y_n = b$ then $|x_n-b|
Let $x_n, y_n \in \mathbb{R}$, $\lim x_n=a$, $\lim y_n = b$ then $|x_n-b|<r<|y_n-a|$ for all $n\in\mathbb N \implies r = |a-b|$.
What I intend to do is to show that $|a-b|-\epsilon<r<|a-b|+\epsilon, \forall \epsilon>0$.
For all $n>N_0$ we have,
$$|y_n-a| = |y_n-b+b-a| \leq |y_n-b|+|b-a| \leq \epsilon+|b-a| $$ 
Then $r\leq \epsilon +|b-a|$. Now,
\begin{align*}
|b-a|-\epsilon &= |b-y_n+y_n-a|-\epsilon\\
&\leq |b-y_n|+|y_n-a|-\epsilon\\
&\leq \epsilon+|y_n-a|-\epsilon = |y_n-a|<r
\end{align*}
$|b-a|-\epsilon<r<|b-a|+\epsilon \implies |r-|b-a||<\epsilon\implies r = |b-a|\quad\quad\quad \blacksquare$
Is it correct?

per Squeeze Theorem if $f(x),g(x),h(x)$ are functions defined on an interval at all points except possibly at $p$ and $f(x)\le g(x)\le h(x)$ for every $x$ not equal to $p$ and that 
$$\lim_{x\to p}f(x)=\lim_{x\to p}h(x)=L$$
then $$\lim_{x\to p}g(x)=L$$
equivalently if they are sequences $f_n,g_n,h_n$ then $$\lim_{n\to p}f_n=\lim_{n\to p}h_n=L\Longrightarrow \lim_{n\to p}g_n=L$$
under the same conditions.
we have for $x_n, y_n \in \mathbb{R}$,$\space n\in\mathbb{N}$ $$\lim_{n\to \infty}x_n=a,\lim_{n\to \infty}y_n = b$$ and $$\forall_{n} |x_n-b|<r<|y_n-a|$$
so for a specific but arbitrarily chosen $n\in\mathbb{N}$
$|x_n-b|<r<|y_n-a|\Longleftrightarrow (|x_n-b|<r)\wedge (r<|y_n-a|)$
$|x_n-b|<r\Longrightarrow (|x_n-b|<r)\vee (|x_n-b|=r)\Longleftrightarrow |x_n-b|\le r,\space$by disjunctive amplification
$|x_n-a|>r\Longrightarrow (|x_n-a|>r)\vee (|x_n-a|=r)\Longleftrightarrow |x_n-a|\ge r,\space$by disjunctive amplification
so we have $|x_n-b|\le r\le |y_n-a|$
let sequences $q_n=|x_n-b|,v_n=|y_n-a|$
then
$$\lim_{n\to \infty}q_n=|a-b|=\lim_{n\to \infty}v_n=|b-a|=|a-b|$$
so by the Squeeze Theorem
$$\lim_{n\to\infty}r=r=|a-b|$$
and so $r=|a-b|$  
