I'm working on my bachelor thesis, and I'm studying principally on two textbooks (Selected Topics on Analysis in Metric Spaces [1] by Luigi Ambrosio and Paolo Tilli and A Course in Metric Geometry [2] by Dmitri Burago, Yuri Burago and Sergei Ivanov) but neither of them reports the detail which I need.
In particular [1] gives this definition:
Definition 4.5.1 (Gromov distance). Let $X$, $Y$ be compact and nonempty metric spaces. We define $d_{GH} (X,Y)$ as the infimum of all $\epsilon >0$ such that there exist a compact metric space $Z$ and isometric embeddings $i_X : X \to Z$, $i_Y : Y \to Z$ such that $d_H (i_X (X) , i_Y (Y)) < \epsilon$ ( - here $d_H$ is the Hausdorff metric).
Then
Proposition 4.5.1. $d_{GH}$ is a metric in the isometry class $\mathcal{M}$ of nonempty compact metric space.
Proof. [...] Now we prove that $d_{GH} (X,Y)=0$ implies that $X$ and $Y$ are isometric. We first notice that if $X$ and $Y$ are contained in the same metric space $Z$, the map $j:X \to Y$ which associates to $x \in X$ any point $y \in Y$ of least distance from $x$ satisfies $$d_Z (x_1 , x_2) - 2 \delta \le d_Z (j(x_1),j(x_2)) \le d_Z (x_1, x_2 ) + 2 \delta \quad \forall x_1 , x_2 \in X \quad (*)$$ provided $d_{GH} (X,Y) < \delta$. Composing with isometries, also in general case we have a map $j: X \to Y$ such that $(*)$ holds, just replacing $d_Z$ with $d_Y$ and $d_X$. Thus, if $d_{GH}(X,Y)=0$ we can find a sequence of maps $j_h : X \to Y$ such that $$d_X (x_1 , x_2) - 2^{-h} \le d_Y (j(x_1),j(x_2)) \le d_X (x_1, x_2 ) + 2^{-h} \quad \forall x_1 , x_2 \in X \quad (**) $$Let $D \subset X$ be a dense set; possibly extracting a subsequence, by a diagonal argument we can assume that $j_h (x)$ converges for any $x \in D$. [...]
My problem is with that diagonal argument: I cannot understand how to precisely construct this functions sequence (I've tried by myself, but I got nowhere); it seems to me that authors, at least a bit, are throwing smoke in reader's eyes (the same happens in [2]).
Thank you in advance.
I'm sorry for my absence. Anyway, the answer I've given is the following one: in the notation above, since we are working in a compact, the sequence $\{j_n (x_1) \}_{n \ge 1}$ admits a convergent subsequence $\{j_{n_i} (x_1) \}_{n_i \in I_1}$ with $| I_1 | = \aleph_0$; we call $j(x_1) \in Y$ the limit of this first subsequence. Similarly we can pick an $I_2 \subseteq I_1$ such that $\{j_{n_i}(x_2) \}_{n_i \in I_2}$ is convergent to $j(x_2)$. At the k-th step we have $I_k \subseteq I_{k-1} \subseteq \dots \subseteq I_2 \subseteq I_1$ and $\{j_{n_i}(x_k) \}_{n_i \in I_k}$ such that $j_{n_i} (x_k) \to j(x_k)$; at this point it is sufficient to take $\{n_j \}_{j \ge 1}$ with $n_j \in I_j$ and $n_j < n_{j+1}$, and then we have that $j_n (x_k)$ converges $\forall x_k \in D$.