If $i[X]$ where not dense in $(\tilde{X}, \tilde{d})$, pick $p$ in $\tilde{X}$ not in the closure of $i[X]$. We know $r>0$ exists so that $B_{\tilde{d}}(p,r) \cap i[X] = \emptyset$.
Define $Y=\overline{i[X]}$ in the restricted metric $d_Y$ from $\tilde{d}$, and note it is a complete metric space (being closed in one) and $f=i$ is well-defined isometry from $(X,d)$ to $(Y,d_Y)$. Now suppose an isometry $g: (\tilde{X}, \tilde{d}) \to (Y,d_Y)$ completing the diagram (so $g \circ i = f$) existed, then where can $p$ go?
$g(p) \in Y$ so lies (in $\tilde{X}$) in the closure of $i[X]$, so for some $x_0 \in X$, $\tilde{d}(i(x_0), g(p)) < r$.
But $i(x_0)= f(x_0) = g(i(x_0))$ and then $d_Y(g(p), g(i(x_0)) < r$ while we knew $\tilde{d}(p, i(x_0)) \ge r$, so $g$ is not an isometry.
So $i[X]$ has to be dense in $\tilde{X}$.
Sometimes you just have to deal with more complicated notation. Here, I shall use a capital letter like $A$ to denote an element of $X^*$; and since $A$ is an equivalence class of sequences in $X$, I shall use something like $(a_n)$ to denote an element of $A$.
Here are some preliminary observations which will help to simplify the proof of completeness:
If $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence in $X$ then every subsequence $(a_{n_k})_{k=1}^{\infty}$ is related via $\sim$ to the main sequence (I leave it to you to verify).
For any $\eta>0$, there is an $N\in \Bbb{N}$ such that for all $m,n\geq N$, we have $d(a_n,a_m) \leq \eta$. In other words, the subsequence $\{a_N, a_{N+1}, \dots\}$ has the property that each pair of terms is at most $\eta$ away from each other. To summarize: for every $\eta>0$, there is a subsequence $(a_{n_k})_{k=1}^{\infty}$ such that for all $k,l\in \Bbb{N}$, we have $d(a_{n_k}, a_{n_l})< \eta$.
Fix a sequence $(\zeta_n)$ of positive numbers which decrease to zero (such as $\zeta_n = \frac{1}{n}$). Now, to show completeness of $X^*$, we have to show every Cauchy sequence converges. So, let $(A_n)_{n=1}^{\infty}$ be a Cauchy sequence in $X^*$. For each $n\in \Bbb{N}$, choose a representative $(a_{n,k})_{k=1}^{\infty} \in A_n$ such that for all $k,l\in \Bbb{N}$, we have
\begin{align}
d(a_{n,k}, a_{n,l})< \zeta_n \tag{i}
\end{align}
Note that such a representative always exists by our remarks above.
Since $(A_n)$ is Cauchy, for every $j\in \Bbb{N}$, there is an $N_j\in \Bbb{N}$ such that for all $n,m,k \geq N_j$, we have
\begin{align}
d(a_{n,k}, a_{m,k}) < \zeta_j \tag{ii}
\end{align}
(just unwind the definition of $(A_n)$ being Cauchy and the definition of $\delta$ to see why this follows). Also observe that by doing this recursively, then you can arrange for this such that $j<N_j$ and $N_1< N_2< N_3\dots$
Now, put $\beta_j:= a_{N_j, N_j}$. We claim that $(\beta_j)_{j=1}^{\infty}$ is Cauchy in $X$. This is because for every $l\in \Bbb{N}$, if $i,j\geq l$ then (since $N_i,N_j \geq N_l$)
\begin{align}
d(\beta_i, \beta_j) &:= d(a_{N_i, N_i}, a_{N_j, N_j}) \\
&\leq d(a_{N_i, N_i}, a_{N_i, N_j}) + d(a_{N_i, N_j}, a_{N_j, N_j}) \\
&\leq \zeta_i + \zeta_l \tag{by i and ii} \\
&\leq 2\zeta_l,
\end{align}
where the last line is because we chose the $\zeta$ sequence to be decreasing. As $l\to \infty$, the RHS tends to $0$, which proves the sequence is Cauchy in $X$.
Finally, let $B:= [(\beta_j)_{j=1}^{\infty}]$ be the equivalence class; I leave it to you to show $A_n \to B$.
Note that the idea of the proof is pretty simple. We have a Cauchy sequence $(A_n)$. We then choose representatives $(a_{n,k})_{k=1}^{\infty}$. So if we write this out as a square array of numbers:
\begin{align}
\begin{matrix}
A_1: & a_{1,1} & a_{1,2} & a_{1,3} & \cdots \\
A_2: & a_{2,1} & a_{2,2} & a_{2,3} & \cdots \\
A_3: & a_{3,1} & a_{3,2} & a_{3,3} & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}
\end{align}
($n$ is going down, $k$ is going to the right). Very informally, (i) says that if you go vertically down far enough, then all the elements in that row will be close to each other. (ii) says that if you go to the far enough to the "bottom right" then all the elements in the same column will be close enough.
So, the idea is to take the diagonal elements $\beta_j := a_{N_j,N_j}$, and show that this has the desired properties. To really understand the proof, I would highly recommend you to write your own arguments for why $(\beta_j)$ is Cauchy and why $A_n \to B$; use the square array above as your guiding principle to see which elements are close to which ones.
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