Proof, that $d(x_n,y_n)$ converges to $d(x,y)$(proof explanation) In a metric space $(S,d)$, assume that $x_n\,\to\,x$ and $y_n\,\to\,y$. Prove that $d\left(x_n,y_n\right)\,\to\,d(x,y)$.
$\textbf{Proof:}$ Since $x_n\,\to\,x$ and $y_n\,\to\,y$, given $\epsilon>0$ there exists a positive integer $N$ such that as $n\ge N$, we have
\begin{equation}
d\left(x_n,x\right)<\epsilon/2 \text{ and }d\left(y_n,y\right)<\epsilon/2
\end{equation}
Hence, as $n\ge N$, we have
\begin{eqnarray}
\left\lvert\,d\left(x_n,y_n\right)-d\left(x,y\right)\,\right\rvert&\le&\left\lvert\,d\left(x_n,x\right)+d\left(y_n,y\right)\,\right\rvert\\
&=&d\left(x_n,x\right)+d\left(y_n,y\right)\\
&<&\epsilon/2+\epsilon/2\\
&=&\epsilon
\end{eqnarray}
I cannot understand why this is true:
\begin{equation}
\left\lvert\,d\left(x_n,y_n\right)-d\left(x,y\right)\,\right\rvert\le\left\lvert\,d\left(x_n,x\right)+d\left(y_n,y\right)\,\right\rvert
\end{equation}
Can you please explain me that?
 A: Hint:
By the triangle inequality,
$$ d(x,y) \leq d(x,u) + d(u,v) +d(v,y) $$
A: By the triangle inequality,
$$d(x_n,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n)\tag 1$$ and
$$d(x,y)\leq d(x,x_n)+d(x_n,y_n)+d(y_n,y)\tag 2$$
From $(1)$, we get 
$$d(x_n,y_n)-d(x,y)\leq d(x_n,x)+d(y_n,y)\tag 3$$ 
and from $(2)$, we get
$$-[d(x_n,x)+d(y_n,y)]\leq d(x_n,y_n)-d(x,y)\tag 4$$
Combining $(4)$ and $(3)$, we get
$$-[d(x_n,x)+d(y_n,y)]\leq d(x_n,y_n)-d(x,y)\leq d(x_n,x)+d(y_n,y)$$
and since $d(x_n,x)+d(y_n,y)\geq 0$, we get 
$$\Big|d(x_n,y_n)-d(x,y)\Big|\leq d(x_n,x)+d(y_n,y).$$
A: Temporarily assume $d(x_n,y_n) \geq d(x,y)$.
Then using the triangle inequality for each step of this reasoning we have:
$$
d(x_n, y) \leq d(x_n,x) + d(x,y)\\
d(x_n,y_n) \leq d(x_n, y) + d(y,y_n) \leq d(x_n,x) + d(x,y) + d(y,y_n) 
$$
And subtracting $d(x,y)$ from both sides
$$
|d(x_n,y_n) -d(x,y)| = d(x_n,y_n) -d(x,y)  \leq d(x_n,x)  + d(y,y_n) \leq |d(x_n,x)  + d(y,y_n)|
$$
So the statement holds if $d(x_n,y_n) \geq d(x,y)$.
The same type of reasoning (about going around a quadrilateral in two ways) holds when  $d(x_n,y_n) < d(x,y)$. In fact, if you just swap $x_n \leftrightarrow x$ and $y_n \leftrightarrow y$ in the above proof, it works for 
$d(x_n,y_n) \leq d(x,y)$.
