In the following book, page 338, there is Lemma A.23:
If $(X,d)$ is a complete metric space, then $(\mathrm{Cpt}(X),d_H)$ is a complete metric space.
I want to understand some details:
(i) Why is $K_n \subseteq \bigcup_{i=1}^{L} B(x_i, 3\epsilon)$? I think I proved with the triangle inequality that $K_n \subseteq \bigcup_{i=1}^{L} B(x_i, 2\epsilon)$ by the following: Let $x \in K_n$, then $x \in K_N^{\epsilon}$, so there is $y \in K_N$ such that $d(x,y)<\epsilon$; Since $K_N$ is covered by balls of radius $\epsilon$, it is true $d(x,x_i) \le d(x,y) + d(y,x_i) < 2\epsilon$. So why would Bishop-Peres choose $3\epsilon$? There must be something wrong with my reasoning.
(ii) I'm also failing to understand why $K \subseteq \bigcup_{i=1}^{L} B(x_i,4\epsilon)$, and probably for a similar reason; I don't see from where this $4$ comes from.
Notation: $\mathrm{Cpt}(X)$ is the set of all nonempty compact subsets of $X$; $K_N^{\epsilon}$ is the set of all $x \in X$ such that $d(x,K_N)< \epsilon$; $d_H$ is the Hausdorff distance.
You are correct in (i). You used only that $x\in A^\epsilon$ implies that $d(x, y) <\epsilon$ for some $y\in A$: this is true even if $A$ is non-compact. I can see that if the initial condition was $d_H(K_n, K_N) \le \epsilon$ for all $n\ge N$, then more care has to be taken; but it is clear that they are using $d_H( K_n, K_N)<\epsilon$.
For (ii), we do need to enlarge a bit:
We already know that $$ K_n \subset \bigcup _i B(x_i, 3\epsilon),\ \ \text{ for all } n\ge N.$$
Then $$ \overline{\bigcup_{j > n} K_j} \subset \overline{\bigcup _i B(x_i, 3\epsilon)}$$ for all $n\ge N$. Hence
$$K := \bigcap_n \overline{\bigcup_{j > n} K_j} \subset \overline{\bigcup _i B(x_i, 3\epsilon)}$$ and thus they have to enlarge the balls to $B(x_i, 4\epsilon)$ because of the closure.