Here is Prob. 24, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Let $X$ be a metric space.
(a) Call two Cauchy sequences $\left\{ p_n \right\}$, $\left\{ q_n \right\}$ in $X$ equivalent if $$ \lim_{n \to \infty} d \left( p_n, q_n \right) = 0.$$ Prove that this is an equivalence relation. [ This is easy.]
(b) Let $X^*$ be the set of all equivalence classes so obtained. If $P \in X^*$, $Q \in X^*$, $\left\{ p_n \right\} \in P$, $\left\{ q_n \right\} \in Q$, define $$ \Delta (P, Q) = \lim_{n \to \infty} d \left( p_n, q_n \right); $$ by Exercise 23, this limit exists. Show that the number $\Delta (P, Q)$ is unchanged if $\left\{ p_n \right\}$ and $\left\{ q_n \right\}$ are replaced by equivalent sequences, and hence that $\Delta$ is a distance function in $X^*$. [ I don't have any problem with that either.]
(c) Prove that the resulting metric space $X^*$ is complete. [ How to do this? ]
(d) For each $p \in X$, there is a Cauchy sequence all of whose terms are $p$; let $P_p$ be the element of $X^*$ which contains this sequence. Prove that $$ \Delta \left( P_p, P_q \right) = d(p, q) $$ for all $p, q \in X$. In other words, the mapping $\varphi$ defined by $\varphi (p) = P_p$ is an isometry (i.e. a distance-preserving mapping) of $X$ into $X^*$. [ This isn't a problem either. ]
(e) Prove that $\varphi(X)$ is dense in $X^*$, and that $\varphi(X) = X^*$ if $X$ is complete. [ How to? ] By (d), we may identify $X$ and $\varphi(X)$ and thus regard $X$ as embedded in the complete metric space $X^*$. We call $X^*$ the completion of $X$.
My effort:
Prob. 24 (c):
Let $\left\{ P_k \right\}$ be a Cauchy sequence in $X^*$. Then, given any $\varepsilon > 0$, we can find a natural number $K$ such that $$\Delta \left( P_k, P_r \right) < \varepsilon$$ for all $k, r \in \mathbb{N}$ such that $k > K$ and $r > K$.
For each $k \in \mathbb{N}$, let $\left\{ p_{kn} \right\}$ be an element of $P_k$. Then, for each $k \in \mathbb{N}$, the sequence $\left\{ p_{kn} \right\}$ is a Cauchy sequence in $(X, d)$ and also $$ \lim_{n \to \infty } d \left( p_{kn}, p_{rn} \right) < \varepsilon$$ for all $k, r \in \mathbb{N}$ such that $k > K$ and $r > K$.
What next?
Prob. 24 (e):
Let $P \in X^*$ and let $\varepsilon > 0$ be given. We need to find an element $x \in X$ such that $$ \Delta (P, \varphi(x) ) < \varepsilon.$$ Let $\left\{ p_n \right\}$ be a Cauchy sequence in $P$. Then there exists a natural number $N$ such that $$ d \left( p_m , p_n \right) < \frac{\varepsilon}{2} $$ for all $m, n \in \mathbb{N}$ such that $m > N$ and $n > N$.
Let $P_{p_{N+1}}$ be the element of $X^*$ that contains the Cauchy sequence $$ \left\{ p_{N+1}, p_{N+1}, p_{N+1}, \ldots \right\}.$$ Then we see that $$ \Delta \left( P, P_{p_{N+1}} \right) = \lim_{n \to \infty} d \left( p_n, p_{N+1} \right) \leq \frac{\varepsilon}{2} < \varepsilon. $$
And, $P_{p_{N+1}} = \varphi(p_{N+1})$. Is this argument correct? If not, then where lies the flaw? What it the correct proof?
An afterthought:
Here's a proof of Prob. 24 (c) that I propose.
Let $\left\{ P_k \right\}_{k \in \mathbb{N}}$ be a Cauchy sequence in $X^*$. Then, given a real number $\varepsilon > 0$, we can find a natural number $K$ such that $$ \Delta \left( P_k, P_r \right) < \frac{\varepsilon}{2}$$ for all $r, k \in \mathbb{N}$ such that $r > K$ and $k > K$.
For each $k \in \mathbb{N}$, since $P_k \in X^*$, each $P_k$ is an equivalence class of Cauchy sequences in $X$. Let $\left\{ p_{kn} \right\}_{n \in \mathbb{N}}$ be a Cauchy sequence in the equivalence class $P_k$, for each $k \in \mathbb{N}$. Then we can conclude that $$ \Delta \left( P_k, P_r \right) = \lim_{n \to \infty} d \left( p_{kn}, p_{rn} \right) < \frac{\varepsilon}{2}$$ for all $r, k \in \mathbb{N}$ such that $r > K$ and $k > K$.
Therefore we can find a natural number $N$ such that $$ d \left( p_{kn}, p_{rn} \right) < \frac{\varepsilon}{2}$$ for all $r, k, n \in \mathbb{N}$ such that $r > K$, $k > K$, and $n > N$. So $$ d \left( p_{kn}, p_{K+1, \ N+1} \right) < \frac{\varepsilon}{2}$$ for all $k, n \in \mathbb{N}$ such that $k > K$ and $n > N$.
Now $$\left\{ p_{K+1, \ N+1}, \ p_{K+1, \ N+1}, \ p_{K+1, \ N+1}, \ \ldots \right\}$$ is a Cauchy sequence in $X$. Let $P \in X^*$ be the equivalence class of this Cauchy sequence.
Now, for each $k \in \mathbb{N}$, since $\left\{ p_{kn} \right\}_{n \in \mathbb{N}}$ is a Cauchy sequence in the equivalence class $P_k$, we can conclude that, for all $k \in \mathbb{N}$ such that $k > K$, the following is true: $$ \Delta \left( P_k, P \right) = \lim_{n \to \infty} d \left( p_{kn}, p_{K+1, \ N+1} \right) \leq \frac{\varepsilon}{2} < \varepsilon.$$ Since $\varepsilon > 0$ was arbitrary, we can conclude that our original sequence $\left\{ P_k \right\}_{k \in \mathbb{N}}$ converges in $\left( X^*, \Delta \right)$ to the point $P \in X^*$. Hence $\left( X^*, \Delta \right)$ is a complete metric space.
Is this proof correct? If not, then where lies the problem? I wonder if the $P\in X^*$ in my argument is dependent on our choice of $\varepsilon$ and is not uniform.