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?
 A: I think your proof is correct. A shorter way of achieving this would be to take limits on both sides of the inequality to get $$|a-b| \le r \le |a-b|$$
(Since $||x_n-b|-|a-b|| \le |x_n-a|$ and $||y_n-a|-|a-b|| \le |y_n-b|$ via Triangle Inequality)
A: No, your calculation has a problem; it is in the very last inequality $\vert y_n -a \vert < r$ in the second chain. 
The correct procedure is as follows: 
We are given that $\left( x_n \right)_{n \in \mathbb{N}}$ and $\left( y_n \right)_{n \in \mathbb{N}}$ are sequences of real numbers and $a$, $b$, and $r$ are real number such that 
$$ \lim_{n\to\infty} x_n = a, \ \tag{1} $$
$$ \lim_{n\to\infty} y_n = b, \ \tag{2} $$
and 
$$ \left\vert x_n - b \right\vert < r < \left\vert y_n - a \right\vert \ \mbox{ for all } n \in \mathbb{N}. \ \tag{3} $$
Let $\varepsilon$ be any positive real number. Then, by the definition of the limit of a sequence, from (1) and (2) we can conclude that, there exists some natural numbers $M$ and $N$ such that 
$$ \left\vert x_n -a \right\vert < \varepsilon \ \mbox{ for all } m \in \left\{ M, M+1, M+2, \ldots \right\}, \ \tag{4} $$
and 
$$ \left\vert y_n - b \right\vert < \varepsilon \ \mbox{ for all } m \in \left\{ N, N+1, N+2, \ldots \right\}. \ \tag{5} $$
Now for any real numbers $\alpha$ and $\beta$, the following chain of inequalities holds:
$$ \left\vert \ \lvert \alpha \rvert -  \lvert \alpha \rvert \ \right\vert \leq \left\vert \alpha \pm \beta \right\vert \leq \left\vert \alpha \right\vert + \left\vert \beta \right\vert. \ \tag{6} $$ 
So for all $n \in \mathbb{N}$ such that $n > \max \left\{ M, N \right\}$, we see that 
$$
\begin{align} 
\left\vert a-b \right\vert &= \left\vert a - x_n + x_n - b \right\vert \\ 
&\leq \left\vert a - x_n \right\vert  +  \left\vert x_n - b \right\vert  \ \mbox{ by (6) above} \\
&< \varepsilon + r, \ \mbox{ by (3) and (4) above} \ \tag{7}
\end{align}
$$ 
and 
$$
\begin{align} 
\left\vert a-b \right\vert &= \left\vert a - y_n + y_n - b \right\vert \\ 
&\geq \left\vert a - y_n \right\vert  -  \left\vert y_n - b \right\vert \ \mbox{ by (6) above } \\
&> r - \varepsilon, \ \mbox{ by (3) and (5) above} \ \tag{8}
\end{align}
$$ 
Thus combining (7) and (8), we conclude that 
$$ r - \varepsilon < \left\vert a-b \right\vert < r + \varepsilon \ \mbox{ for every real number } \varepsilon > 0,$$
which is equivalent to 
$$ \left\vert \ \left\vert a-b \right\vert - r \  \right\vert < \varepsilon \ \mbox{ for every real number } \varepsilon > 0.$$ 
Hence 
$$\left\vert \ \left\vert a-b \right\vert - r \  \right\vert = 0,$$
which implies that 
$$\left\vert a-b \right\vert = r.$$
A: 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 $r=|a-b|$  
