Forward :
Let $x_n$ and $y_n$ be two Cauchy Sequences in $(M,d)$. Also, let $x_n \to L$ and $y_n \to L$ with $L \in M$. This indeed means that :
$$d(x_n,L) \to 0 \quad \text{and} \quad d(y_n,L) \to 0 $$
This can equally be expressed as that $\exists \varepsilon >0$ such that
$$d(x_n,L) < \varepsilon/2 \quad \text{and} \quad d(y_n,L) < \varepsilon/2$$
as $\varepsilon$ can become arbitrary small.
But $d$ is a metric in the space $M$ and thus the Triangle Inequality holds :
$$d(x_n,y_n) \leq d(x_n,L) + d(y_n,L) < \varepsilon \implies d(x_n,y_n) \to 0$$
Reverse :
Let $x_n$ and $y_n$ be two Cauchy Sequences in $(M,d)$, such that $d(x_n,y_n) \to 0$. Now, let $x_n \to x$ and $y_n \to y$ with $x,y \in M$. We are interested to see if their limits coincide, aka if $x$ and $y$ can be brought infinitely close. For this, we will work over their distance under the metric $d$, namingly $d(x,y)$. But then, it is :
$$d(x,y) \leq d(x,x_n) + d(x_n,y_n) + d(y_n,y) $$
Now, note that since $x_n \to x$ it is $d(x_n,x) \to 0$ and the same holds for $d(y_n,y)$. We have also let that $d(x_n,y_n) \to 0$. Thus, it is $d(x,y) \to 0$ and that essentialy means that $x \equiv y$.
Reference/Note : A nice post on why the limit exists.
You can use the quadrilateral inequality:
$$|\rho(a, b) - \rho(c, d)| \le \rho(b, c) + \rho(a, d).$$
This inequality can be proven using triangle inequality. This inequality gives us
$$|\rho(x_n, y_n) - \rho(x_m, y_m)| \le \rho(x_n, x_m) + \rho(y_n, y_m),$$
where the latter two can be made arbitrarily small.
Best Answer
Let $\varepsilon > 0.$
Since we know that $x_n \to L$, we know that there exists $N_x$ such that when $n>N_x,$ then $d(x_n,L)<\frac{\varepsilon}{2}$.
Likewise, since $y_n \to L$, we know that there exists $N_y$ such that when $n>N_y$, $ $ $d(y_n,L)< \frac{\varepsilon}{2}.$
Set $N =$max$\{N_x,N_y\}$.
Then when $n >N,$ by the Triangle Inequality: $$d(x_n,y_n) \leq d(x_n,L) + d(y_n,L) = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$
Thus, we have shown that if $x_n \to L$ and $y_n \to L,$ $d(x_n,y_n) \to 0.$ $\square$