In a metric space $(S,d)$, assume that $x_n \to x$ and $y_n \to y$. Prove that $d(x_n, y_n) \to d (x, y)$. Let $(X,d)$ be a metric space and $(x_n)$,$(y_n)$ be sequences in $X$.
(a) If $x_n \to x$ and $y_n \to y$, prove that $d(x_n,y_n) \to d(x,y)$.
(b) If $x_n$ and $y_n$ are Cauchy sequences in $X$, prove that the real sequence $d(x_n,y_n)$ is convergent.
Proof. (a) Is this correct?: Let $x,y,w,z∈X$. The triangle inequality implies that $|d(x,z)-d(w,y)| \leq d(x,y)+d(z,w)$, so 
$|d(x_n,y_n)-d(x,y)| \leq d(x_n,x)+d(y_n,y)$
which implies that $d(x_n,y_n) \to d(x, y)$ as $n \to ∞$ if $d(x_n,x) \to 0$ and
$d (y_n, y) \to 0$.
(b) I have no ideas, you help me please?
 A: Since you seem to have part (a) locked down, I'll answer part (b).
Let $\epsilon \gt 0$ be given. Since $\{x_n\}$ is Cauchy, $\exists  N_1$ such that $\forall n,m \ge N_1, d(x_n,x_m) \lt \frac\epsilon2$. Likewise, since $\{y_n\}$ is Cauchy, $\exists  N_2$ such that $\forall n,m \ge N_2, d(y_n,y_m) \lt \frac\epsilon2$. Take $N = max(N_1, N_2)$.
Then $\forall n,m \ge N$ we have $d(x_n, y_n) \le d(x_n, x_m) + d(x_m, y_m) + d(y_n, y_m)$, so that $d(x_n, y_n) - d(x_m, y_m) \le d(x_n, x_m) + d(y_n, y_m) \lt \frac\epsilon2 + \frac\epsilon2 = \epsilon$.
Similarly, $\forall n,m \ge N$ we have $d(x_m, y_m) \le d(x_m, x_n) + d(x_n, y_n) + d(y_m, y_n)$, so that $d(x_m, y_m) - d(x_n, y_n) \le d(x_m, x_n) + d(y_m, y_n) \lt \frac\epsilon2 + \frac\epsilon2 = \epsilon$.
Thus we have shown that $\forall \epsilon \gt 0 \exists N$ such that $d(d(x_n,y_n),d(x_m,y_m)) \lt \epsilon$, so that the sequence $\{d(x_n,y_n)\}$ is Cauchy and hence convergent. 
A: Your proof of a looks good.
For b, it's the same idea as for a, but you don't have access to $x$ and $y$. Therefore, you need to use $d(x_n,x_m)$ and $d(y_m,y_m)$ for sufficiently large $n$ and $m$ to show that the sequence $d(x_n,y_n)$ is Cauchy, and therefore convergent.
A: In part (a), your proof is correct. I would include a proof of the key observation, namely that

The triangle inequality implies that $|d(x,z)-d(w,y)| \le d(x,y)+d(z,w)$.

For part (b), let $(X',d')$ be the completion of $(X,d)$. Now you have that the Cauchy sequences $x_n$ and $y_n$ converge, and so part (a) immediately gives you the result.
