Baby Rudin Ex 3.24 
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.
(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 $X^*$.
(c) Prove that the resulting metric space $X^*$ is complete.

I am referring to the solution manual by Roger Cooke. Parts (a) and (b) were quite easy. I couldn't quite comprehend the solution of (c). Presented below is my argument done in a slightly different manner :
Let $(P_k)$ be a Cauchy sequence in $X^*.$ Denote $(p_{kn})_{n\in \mathbb{N}}$ as the Cauchy sequence in $X$ representing $P_k.$
Thus for any $\epsilon >0, \,\exists K_{\epsilon}\in \mathbb{N}$ s.t $$\Delta(P_i,P_j)<\epsilon, \hspace{1cm}\forall i,j\ge K_{\epsilon}.$$
This implies there exists $N_{\epsilon}\in \mathbb{N}$ s.t $$d(p_{in},p_{jm})<\epsilon, \hspace{1cm}\forall i,j\ge K_{\epsilon},\forall m,n\ge N_{\epsilon}.$$
Let $\epsilon_n = 1/n.$ Correspondingly for every such $\epsilon_n$ we get $K_{\epsilon_n}$ and $N_{\epsilon_n}.$ Choose subsequences $K_n$ and $N_n$ in an increasing fashion.
Thus if $ i,j\ge K_{n_0}\&\, m,n\ge N_{n_0}$ then $$d(p_{in},p_{jm})<\frac{1}{n_0}. $$
Define a sequence $(p_n)$ in $X$ such that $p_n :=p_{K_nN_n}$
Claim : $(p_n)$ is Cauchy.
We observe that for any $\epsilon >0$ there exists some $N_0\in \mathbb{N}$ such that $\frac{2}{N_0}<\epsilon$ (Archimedean Property). Then if $m>n\ge N_0$ we have $K_m>K_n>K_{1/{N_0}}$ and $N_m>N_n>N_{1/{N_0}}$ and thus
\begin{align}
d(p_m,p_n)=d(p_{K_mN_m},p_{K_nN_n})\le d(p_{K_mN_m},p_{K_{N_0}N_{N_0}})+d(p_{K_{N_0}N_{N_0}},p_{K_nN_n})< \frac{1}{N_0}+\frac{1}{N_0} =\frac{2}{N_0}<\epsilon.
\end{align}
Thus $(p_n)$ is Cauchy.
Let $P\in X^*$ be the associated element of this sequence.
Claim : $(P_k)$ converges to $P.$
It is to be shown that for any $\epsilon>0$ there exists $K\in \mathbb{N}$ such that $$\Delta(P_k,P)=\lim_{n\to \infty} d(p_{kn},p_n)<\epsilon,\hspace{0.5cm}\forall k\ge K.$$
Let $N_0\in \mathbb{N}$ be such that $\frac{1}{N_0}<\epsilon.$ Then for any $k\ge K_{N_0}$ and $n\ge N_{N_0}$ we have
$$d(p_{kn},p_n)<\frac{1}{N_0}<\epsilon, \hspace{0.5cm} \forall k\ge K_{N_0},n\ge N_{N_0}.$$
This implies $$\lim_{n\to \infty} d(p_{kn},p_n)<\epsilon,\hspace{0.5cm}\forall k\ge K_{N_0}$$
i.e., $\forall \epsilon >0,\,\exists K\in \mathbb{N}$ such that
$$\Delta(P_k,P)<\epsilon,\hspace{0.5cm}\forall k\ge K$$
Thus $(P_k)$ converges to $P.\hspace{7cm} \blacksquare$

Is my argument correct?

 A: You state the following:

This implies there exists $ N_{\epsilon} \in \mathbb{N}$ s.t. 
  $$ d(p_{in},p_{jm})<\epsilon, \hspace{1cm}\forall i,j\ge K_{\epsilon},\forall m,n\ge N_{\epsilon}. $$

It is true that for every individual pair of indices $i,j$ (and corresponding sequences) there exists such an $N_\epsilon$, but what is the reasoning behind the statement saying that there is one $N_\epsilon$ that can be applied to every pair of sequences for which $i,j \ge K_\epsilon$ holds? If there were only finite number of such sequences, you could take maximum of all individual pairwise $N_\epsilon$ values and use that. But in this case there are infinite (countable) number of sequences, so taking maximum does not work (the maximum might not even exist).
For example, one could construct real sequences such that 
$$ p_{kn} =
\begin{cases}
1,    & \text{if } n \lt k \\\\
2^{-n+k}, & \text{otherwise}
\end{cases}
$$
Now the sequence $p_{1n}$ would begin with $1,\frac{1}{2}, \frac{1}{4}, ...$, $p_{2n}$ would begin with $1, 1,\frac{1}{2}, \frac{1}{4}, ...$, $p_{3n}$ would begin with $1, 1, 1,\frac{1}{2}, \frac{1}{4}, ...$, and so forth. So sequences $(p_{kn})$ would have an increasing amount of leading ones, but still every sequence would converge to $0$. Thus they are also Cauchy sequences and if equivalence classes in Rudin's exercise 3.24 are considered, these sequences would even belong to the same equivalence class. Now for every $\epsilon > 0$ and every pair of sequences you can find the $N_\epsilon$ such that $n,m \ge N_\epsilon$ implies $d(p_{in},p_{jm}) < \epsilon$. But still you cannot find any single $N_\epsilon$ that would imply the same inequality with every sequence! (This is easy to see by contradiction: Just take an epsilon less than one (say $1/2$, for example) and assume such an $N_\epsilon$ exists. Then you can always choose a sequence $(p_{kn})$ where $k > N_\epsilon$ and you can find the contradiction.)
