Limits coincide distance tends to zero? $\newcommand{\N}{\mathbb{N}}$
$\newcommand{\max}{\operatorname{Max}}$

Let $(X,d)$, $X \not= \emptyset$, a
metric space and $(x_{n})_{n \in \N}$ and $(y_{n})_{n \in \N}$
sequences in $X$. Show that if there exists a $a \in X$ such that:
$$\lim_{n \to \infty}{x_{n}} = a = \lim_{n \to \infty}{y_{n}}$$
Then:
$$\lim_{n \to  \infty}{d(x_{n},y_{n})} = 0$$
Is the converse also true? (proof or counterexample).

Here is my attempt:
Suppose that:
$$\lim_{n \to \infty}{x_{n}} = a = \lim_{n \to \infty}{y_{n}}$$
We want to see that:
$$\lim_{n \to  \infty}{d(x_{n},y_{n})} = 0$$
Let $\varepsilon > 0$, take $\varepsilon = \frac{\varepsilon}{2}$ so in both cases there exists $N_{1},N_{2} > 0$ such that:
$$\forall_{n > N_{1}}\left[ d(x_{n},a) < \frac{\varepsilon}{2}\right]$$
$$\forall_{n > N_{2}}\left[ d(y_{n},a) < \frac{\varepsilon}{2}\right]$$
Now, set $N = \max{N_{1},N_{2}}$. Then, if we add both equations we have that:
$$\forall_{n > N} [d(x_{n},a) + d(y_{n},a) < \varepsilon]$$
Using the triangular inequality:
$$\forall_{n > N} [d(x_{n},y_{n}) < \varepsilon]$$
In particular:
$$\forall_{n > N} [|d(x_{n},y_{n}) - 0 |< \varepsilon ]$$
In fact we find that:
$$\forall_{\varepsilon}\{\exists_{N > 0} \{\forall_{n > N_{1}}[ |d(x_{n},y_{n}) - 0| <\varepsilon]\}\}$$
So by definitiuon of limit:
$$\lim_{n \to \infty}{d(x_{n},y_{n})} = 0$$
What do you think?
 A: Let $\varepsilon>0$. Let $N_1 \in \Bbb N$ be such that
$$\forall n \ge N_1: d(x_n, a) < \frac{\varepsilon}{2}\tag{1}$$
by $\displaystyle \lim_{n \to \infty} x_n =a$ and $N_2 \in \Bbb N$ such that
$$\forall n \ge N_2: d(y_n, a) < \frac{\varepsilon}{2}\tag{2}$$
by $\displaystyle \lim_{n \to \infty} y_n =a$. Now set $N=\max(N_1,N_2)$ and use
$$\forall n: d(x_n,y_n) \le d(x_n,a) + d(y_n, a)$$
to finish showing that
$$\forall n \ge N: |d(x_n, y_n) - 0| < \varepsilon$$
as is required to show that $\displaystyle \lim_{n \to \infty} d(x_n,y_n) =0$.
The converse need not hold. Let $X$ be the set $\Bbb Q$ and $d(x,y)=|x-y|$ as usual, or even easier $X=\Bbb R\setminus\{0\}$ in that same metric.
A: For the first part $d(x_n,y_n) \leq d(x_n,a)+d(a,y_n)$ can also be used.
The converse holds true only if one of the sequences $(x_n)_n$ and $(y_n)_n$ is Cauchy and the  space $(X,d)$ is complete. Because: from
$d(x_n,x_m) \leq d(x_n,y_n)+d(y_n,y_m)+d(y_m,x_m)$ and $d(y_n,y_m) \leq d(y_n,x_n)+d(x_n,x_m)+d(x_m,y_m)$ we get $\lim_{n,m \rightarrow \infty}d(x_n, x_m)=\lim_{n,m \rightarrow \infty}d(y_n, y_m)$. So $(x_n)_n$ is Cauchy iff $(y_n)_n$ is so. By completeness of $(X,d)$, $\exists~~~a,b \in X$ such that $\lim_{n \rightarrow \infty}d(x_n, a)=\lim_{n \rightarrow \infty}d(y_n, b)=0$. $d(a,b) \leq d(a,x_n)+d(x_n,y_n)+d(y_n,b)$, taking limit as $n \rightarrow \infty$ we have $d(a,b)=0\Rightarrow a=b$.
Example that shows Cauchy condition can't be dropped: take $x_{2n-1}=2n-1$, $x_{2n}=\frac{1}{2n}$ and $y_{2n-1}=2n-1$, $y_{2n}=\frac{1}{n}$ in $(\mathbb{R},|\cdot|)$.
That the completeness condition can't be dropped can be seen from @Henno Brandsma's example
