I am trying to solve (d) below:
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.
(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 $X^*$.
My attempt: Note that for any $p, q \in X$, $\Delta \left( P_p, P_q \right) = \lim\limits_{n \to \infty} d(p_n, q_n) = \lim\limits_{n \to \infty} d(p, q)$. Now, how can I show that $\lim\limits_{n \to \infty} d(p, q) = d(p, q)$? Is there an alternative way to prove (d)?
Here I provide an answer to $(a)$ and $(b)$ if you do not have it already.
(a) Let us prove the transitivity property of such relation.
Let $(X,d_{X})$ be a metric space.
Let's also suppose that $a_{n}$ and $b_{n}$ are equivalent as well as $b_{n}$ and $c_{n}$. We shall prove that $a_{n}$ and $c_{n}$ are equivalent.
Let $\varepsilon > 0$. Then there exists $N_{1} \geq 0$ s.t. $d_{X}(a_{n},b_{n}) \leq \varepsilon/2$ whenever $n\geq N_{1}$.
Similarly, there exists $N_{2} \geq 0$ s.t. $d_{X}(b_{n},c_{n}) \leq \varepsilon/2$ whenever $n\geq N_{2}$.
Due to the triangle inequality, if we choose $N = \max\{N_{1},N_{2}\}$, we conclude that \begin{align*} n\geq N \Rightarrow d_{X}(a_{n},c_{n}) \leq d_{X}(a_{n},b_{n}) + d_{X}(b_{n},c_{n}) \leq \varepsilon \end{align*}
Since $d_{X}(a_{n},a_{n}) = 0$ and $d_{X}(a_{n},b_{n}) = d_{X}(b_{n},a_{n})$, the reflexive and symmetric properties are readily proven.
(b) Let $p'_{n}\in P$ and $q'_{n}\in Q$.
On the assumption that $L = \Delta(P,Q) = \displaystyle\lim_{n\to\infty}d_{X}(p_{n},q_{n})$, let us consider the following relation \begin{align*} |d_{X}(p'_{n},q'_{n}) - L| \leq |d_{X}(p'_{n},q'_{n}) - d_{X}(p'_{n},q_{n})| + |d_{X}(p'_{n},q_{n}) - d_{X}(p_{n},q_{n})| + |d_{X}(p_{n},q_{n}) - L| \end{align*}
Now we prove that $\displaystyle\lim_{n\to\infty}d_{X}(p'_{n},q'_{n}) = \lim_{n\to\infty}d_{X}(p'_{n},q_{n})$.
Since $q'_{n}\sim q_{n}$, we have that $\displaystyle\lim_{n\to\infty}d_{X}(q'_{n},q_{n}) = 0$. Hence, due to the triangle inequality, \begin{align*} \begin{cases} d_{X}(p'_{n},q'_{n}) \leq d_{X}(p'_{n},q_{n}) + d_{X}(q_{n},q'_{n})\\\\ d_{X}(p'_{n},q_{n}) \leq d_{X}(p'_{n},q'_{n}) + d_{X}(q'_{n},q_{n}) \end{cases} \end{align*} Taking the limit as $n\to\infty$ it results that $\displaystyle\lim_{n\to\infty}d_{X}(p'_{n},q'_{n}) = \lim_{n\to\infty}d_{X}(p'_{n},q_{n})$.
Analogous reasoning proves that $\displaystyle\lim_{n\to\infty}d_{X}(p'_{n},q_{n}) = \lim_{n\to\infty}d_{X}(p_{n},q_{n})$.
Finally, we have that \begin{align*} \lim_{n\to\infty}|d_{X}(p'_{n},q'_{n}) - L| \leq 0 + 0 + 0 = 0 \Rightarrow \lim_{n\to\infty}d_{X}(p'_{n},q'_{n}) = L \end{align*} and we are done.