Part of the proof that every metric space has a dense completion is to show that the metric $\tilde{d}([(x_n)], [(y_n)]) = \lim_n d(x_n, y_n)$, where $d$ is the original metric (in the possibly incomplete metric space) and $(x_n), (y_n)$ are Cauchy, is independent of choice of representative. I've written my proof up below.
(Note that the equivalence relation I'm using is the standard one: $(z_n) \sim (z'_n)$ (Cauchy sequences) if $\lim_n d(z_n, z'_n) = 0$.)
Proof:
To show it is well defined, we must show that it is independent of choice of representative. Indeed, suppose that $(x_n) \sim (x'_n)$ and $(y_n) \sim (y'_n)$. We must show that $\lim_{n \to \infty} d(x_n, y_n) = \lim_{n \to \infty} d(x'_n, y'_n)$. From the triangle inequality: \begin{align*} d(x_n, y_n) &\leq d(x'_n, y'_n) + d(y'_n, y_n) + d(x_n, x'_n)\\ d(x'_n, y'_n) &\leq d(x_n, y_n) + d(y'_n, y_n) + d(x_n, x'_n). \end{align*} Thus, rearranging the inequalities above gives $|d(x_n, y_n) - d(x'_n, y'_n)| \leq d(y'_n, y_n) + d(x_n, x'_n)$. Passing to the limit superior: $$ \limsup_{n \to \infty} |d(x_n, y_n) - d(x'_n, y'_n)| \leq \lim_n d(y'_n, y_n) + \lim_n d(x_n, x'_n) = 0. $$ But this implies that $\lim_n d(x_n, y_n) = \lim_n d(x'_n, y'_n)$, as desired.
Question (1): Is my proof correct?
Question (2): How do we know that $\lim_n d(x_n, y_n)$ exists? I'm silently relying on this fact in my proof.
First establish the following:
If $(x_n)_n$ and $(y_n)_n$ are $d$-Cauchy, the limit of $d_n:=(d(x_n, y_n))_n$ exists because it is a Cauchy sequence in the reals. Just as in your estimates we get $|d_n -d_m| = |d(x_n, y_n) - d(x_m, y_m)| \le d(x_n, x_m) + d(y_n, y_m)$. And then we use the definition of Cauchy sequence for $\frac{\varepsilon}{2}$ for $(x_n)_n$ and likewise for $(y_n)_n$ and we get that $(d_n)_n$ is Cauchy in the reals, and so convergent (we use that $(\mathbb{R} ,|\cdot|)$ is complete).
By your estimates the real-valued sequence $|d(x_n, y_n) - d(x'_n, y'_n)|$ is upper bounded by a sequence $c_n$ that tends to $0$. So it also tends to $0$, and this implies that the limits of the two sequences $(d(x_n, y_n))_n$ and $(d(x'_n, y'_n))_n$ (which both exist by the first part) are equal.