Given compact metric spaces $X, Y$, a function $f \colon X \to Y$ is an $\varepsilon$-isometry if $f(X)$ is an $\varepsilon$-net in $Y$ (every point in $Y$ is within $\varepsilon$ of a point in $f(X)$), and the distortion of $f$, defined as follows, $$\text{dis}(f) := \sup_{x_{1}, x_{2} \in X}\left|d_{X}(x_{1}, x_{2}) - d_{Y}(f(x_{1}), f(x_{2})) \right|$$ is bounded by $\varepsilon.$ Note that a $\varepsilon$-isometry is not required to be continuous.
Exercise 7.5.8 of "A Course in Metric Geometry" by Burago, Burago, and Ivanov asks the following question (paraphrased here):
Let $X$ be the unit ball in $\mathbb{R}^{n},$ and let $\{X_{k}\}_{k=1}^{\infty}$ be a sequence of compact metric spaces converging to $X$ in the Gromov-Hausdorff sense. Prove that there exists a sequence of continuous maps $f_{k} \colon X_{k} \to X$ such that each $f_{k}$ is an $\varepsilon_{k}$-isometry for a sequence $\varepsilon_{k} \to 0.$
Prove the same is true if $X$ is a sphere $S^{n-1}$, an $n$-torus $T^{n}$, a metric space homeomorphic to any of the previous, a metric space homeomorphic to a compact smooth manifold. Furthermore, prove the same under some weaker (as weak as possible) topological restrictions on $X$.
An answer given here To construct a continuous map on compact metric spaces (cf. Gromov-Hausdorff Limit) uses the following approach: if $d_{GH}(X_{k}, X) < \varepsilon$, then there exists a space $Z$ containing isometrically embedded copies of $X_{k}$ and $X$, such that these copies have Hausdorff distance less than $\varepsilon$.
Then, given a point in $X_{k}$ (identified with its isometric embedding in $Z$), we send it to the closest point in $X$ (also identified with its isometric embedding). It is not difficult to check that this is a $2\varepsilon$-isometry, but there is an issue: if there are multiple points in $X$ equidistant from our point $x,$ how do we pick which one to send $x$ to, and how do we do so continuously?
This is the part of the answer that I am stuck on. The author picks a collection of points $a_{1}, \ldots, a_{n}$ in $X$ (in this case the unit ball), and inductively defines $C_{0}^{x} = \{y \in X \mid d(x, y) = d(x, X)\},$ $d^{x}_{j} = d(a_{j}, C_{j-1}^{x}),$ $C_{j}^{x} = \{y \in X \mid d(a_{j}, y) = d_{j}^{x}\}.$ Essentially, given our collection of points equidistant from $x,$ we take the ones that are closest to $a_{1},$ then we take the ones among those closest to $a_{2},$ and so on for all the $a.$ If we choose our $a_{1}, \ldots, a_{n}$ wisely so that the distances from a point in $X$ to $a_{1}, \ldots, a_{n}$ uniquely determines the point, we end up uniquely determining a point $y$ at the end. Then, we send $x$ to $y.$
What I don't understand is why this is continuous. In particular, why do the $d^{x}_{j}$'s vary continuously with $x$? Let's say you have $x, x'$ in $X_{k}$. We want to say that if $x, x'$ are close, then $d^{x}_{1}$ and $d^{x'}_{1}$ are close. Note that $|d^{x}_{1} - d^{x'}_{1}|$ is bounded above by $$d(C_{0}^{x}, C_{0}^{x'}) = \inf_{z \in C_{0}^{x}, z' \in C_{0}^{x'}} d(z, z').$$ So, if we can show that $x, x'$ close implies $d(C_{0}^{x}, C_{0}^{x'})$ small, then we are done. However, an application of the triangle inequality only gives $$d(C_{0}^{x}, C_{0}^{x'}) \leq d(x, x') + d(x, X) + d(x', X),$$ which doesn't tell us anything because all we know about $d(x, X)$ and $d(x', X)$ is that they are less than $\varepsilon$.
I don't perfectly understand the method used by them but I can answer the specific question asked by observing that it is impossible for a function to be necessarily continuous for all $(X_n, f)$ when defined by $F(x) = S(C_f(x)))$ where $C_f(x) = \{y \in X \mid d(f(x),y) = \epsilon \}$ and $S:\mathcal{P}(X) \rightarrow X$ is a selection function with the property that $S(\{x\}) = x$
Namely if $f:X_n\rightarrow X$ is a discontinuous bijection with distortion $\epsilon$, then $F(x) = f(x)$, and is still discontinuous.
I would recommend instead an approach of first defining continuous functions between $\epsilon$-nets and then making it global with a partition of unity.
This approach also easily generalizes to compact manifolds for the second half of the textbook question. I am not sure how much farther the generality of the theorem can extend and have no counterexample it isn't all compact metric spaces