This is about Gromov Hausdorff limit on compact metric spaces
(Reference A course in metric geometry - Burago Burago and Ivanov, 268p. EXE 7.5.8)
Definition : $d_{GH}(X,Y)<\epsilon $ if there exists a metric space $Z$ s.t. $$ f: X\rightarrow Z,\ g: Y\rightarrow Z$$ are isometries onto its image and $$ d_H(f(X), g(Y))< \epsilon $$ Here $d_H$ is a Hausdorff distance. That is, $\epsilon$-neighborhood of $f(X)$ contains $g(Y)$ and vice versa.
Definition 2 : If $X_n$ is a sequence of compact metric spaces and if $X$ is a compact metric space, then $X_n\rightarrow_{GH}X$ if $$ d_{GH}(X_n,X)\rightarrow 0 $$
Problem : If $(X_n,d_n)\rightarrow (X,d)$, where $X$ is a standard unit ball in ${\bf R}^n$, then there exists continuous map $f_n :X_n\rightarrow X$ s.t. $f_n$ is $\epsilon_n$-isometry and $\epsilon_n\rightarrow 0$ ($\epsilon_n$-isometry : any point in $X$ has distance $<\epsilon_n$ from $f_n(X_n)$ and $\sup_{x,\ y}\ |d_n(x,y)-d(f_n(x),f_n(y)) | <\epsilon_n$)
Proof : For $\delta$, there exists $N$ s.t. $n\geq N$ implies that $d_{GH}(X_n,X)<\delta$
That is for $n$ there exists a metric space $(Z,d_Z)$ s.t. $$ f: X_n\rightarrow Z,\ g: X\rightarrow Z $$ and $$ d_H(f(X_n),g(X))<\delta $$
Here we want to construct $ F : X_n \rightarrow X $ For $z\in Z',\ Z':=f(X_n)\cap g(X)$ and $x=f^{-1}(z),\ g(y)=z$, $F(x)=y$
For $z\in f(X_n)-g(X)$ we want to find a suitable point in $Z'$ For instance if ther exists unique $z'$ s.t. $d(z,Z')=d(z,z'),\ z'\in Z'$, then $F(f^{-1} (z))=g^{-1}(z') $.
And note that $F$ is $\delta$-isometry and continuous.
But this may not happen since $z'$ is not unique So how can we construct ? Thank you in anticipation.
I note from the definition of GH-convergence that $X$ is compact, so we are talking about the closed unit ball. That makes things easier. You cannot guarantee a unique closest element $x' \in f(X)$ to $x$, so you need to find a means to choose between multiple elements.
One way is consider the points $a_k = (\delta_{1k}, \delta_{2k}, \ldots, \delta_{nk})$, where $\delta_{ij}$ is the Kronecker delta. Let $$A_x = \{ z \in f(X) \mid d(x, z) = d(x, f(X))\}$$ If $d_x = d(x, f(X))$, then $A_x = f(X) \cap \{ z \in Z\mid d(z, x) = d_x\}$ is the intersection of a compact set and a closed set, and so must be compact.
Define $C_0 = f^{-1}(A_x)$, then $C_0$ is a closed subset of the compact set $X$, and thus also compact. For each $k, 0 < k \le n$, define $d_k = d(a_k, C_{k-1})$ and
$$C_k = \{ t \in C_{k-1}\mid d(t, a_k) = d_k\}.$$
A similar argument shows that $C_k$ is compact for all $k$. Since $C_{k-1}$ is compact, it contains points $t$ such that $d(t, a_k) = d_k$, so $C_k$ is not empty for any $k$.
Now the points in $C_n$ are all at a distance $d_k$ from $a_k$ for all $k$. If we consider a point $(t_1, ..., t_n)$, then these conditions supply equations $$R + 1 - 2t_k = d_k^2$$ where $R = \sum_k t_k^2$. Solving the equations for $t_k$ and substituting into the expression for $R$ yields a quadratic in $R$ which has at most two solutions. Since $C_n$ is not empty, there must be at least one positive solution. If there are two, we pick the larger. The value of $R$ then completely determines the values of $t_k$, and therefore a unique point $t' \in C_n$. Define $F(x) = f(t')$.