Q:

If $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}y_n=y$ then $\lim_{n\to\infty}d(x_n,y_n)=d(x,y)$

I want to prove the following (Exercise 4.7 Apostol's Mathematical Analysis).

If $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}y_n=y$ then $\lim_{n\to\infty}d(x_n,y_n)=d(x,y)$

Here is my solution. Would you please check that to see if it is OK. Also, any suggestion for improvement is appreciated. Can we make a geometric interpretation for this?
The assumptions imply that
\begin{align*}
&\forall \varepsilon_1>0\,\,\exists N_1\in\mathbb{Z}^+:n\ge N_1\implies d(x,x_n)\lt\varepsilon_1 \\
&\forall \varepsilon_2>0\,\,\exists N_2\in\mathbb{Z}^+:n\ge N_2\implies d(y,y_n)\lt\varepsilon_2
\tag{1}
\end{align*}
and we are to prove
\begin{align*}
\forall \varepsilon>0\,\,\exists N\in\mathbb{Z}^+:n\ge N\implies |d(x_n,n_n)-d(x,y)|\lt\varepsilon.
\tag{2}
\end{align*}
Let us set $N=\max\{N_1,N_2\}$ so the inequalities in $(1)$ hold. By the triangle inequality, we also have
\begin{align*}
|d(x_n,y_n)-d(x,y)|&\le d(x_n,y_n)+d(x,y)\le d(x_n,x)+d(x,y_n)+d(x,y)\\
&\le d(x_n,x)+d(x,y)+d(y,y_n)+d(x,y) \\
&=d(x,x_n)+d(y,y_n)+2d(x,y).
\tag{3}
\end{align*}
Choosing $\varepsilon_1=\varepsilon_2=\frac{\varepsilon}{2}-d(x,y)$, $(1)$ and $(3)$ lead us to
\begin{align*}
|d(x_n,y_n)-d(x,y)|\lt\Big(\frac{\varepsilon}{2}-d(x,y)\Big)+\Big(\frac{\varepsilon}{2}-d(x,y)\Big)+2d(x,y)=\varepsilon
\end{align*}
which completes the proof.

A:

Apply two times the triangular inequality allows to have the same quantity but with reversed sign, thus we get a bound for the absolute value.
$d(x_n,y_n)\le d(x_n,x)+d(x,y)+d(y,y_n)$
$d(x,y)\le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$
This gives $\bigg|d(x,y)-d(x_n,y_n)\bigg|\le d(x_n,x)+d(y_n,y)\le \epsilon_1+\epsilon_2$

