In a metric space $(S,d)$, assume that $x_n \to x$ and $y_n \to y$. Prove that $d(x_n, y_n) \to d (x, y)$.

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.