I'm working on proving that every compact metric space is the continuous image of the Cantor set (we use the definition of the Cantor set being a set homeomorphic to $\{0,1\}^{\mathbb{N}}$). The question of proving this has been asked several times here on stackexchange, so hopefully this doesn't get flagged; my question is more on justifying a technical point.
Attempt
I understand the general form (roughly) of the proof is as follows: Take the compact set $X$ and cover it with two closed balls $U_{0},U_{1}$. Then take $U_1$ and $U_2$ and cover each with two closed balls $U_{0,0},U_{0,1}$, and $U_{1,0}$, $U_{1,1}$, respectively. By replacing $U_{0,0}$ and $U_{0,1}$ with $U_{0}\cap U_{0,0}$ and $U_{0}\cap U_{0,1}$, respectively, we can assume that $U_{0,0}$ and $U_{0,1}$ are contained in $U_{0}$; for the same reasons we can also assume that $U_{1,0},U_{1,1}$ are contained in $U_{1}$. Continuing this process inductively yields sets $$\mathscr{U}:=\big\{U_{t}\,:\,t\in\{0,1\}^{\mathbb{N}}\big\}.$$ Now each set $U_t$ is essentially constructed as the infinite intersection of nested compact sets, so if I could further rig up the construction to have these nested sets decrease to zero in diameter, each $U_{t}\in\mathscr{U}$ is a singleton, so the function $f:\{0,1\}^{\mathbb{N}}\to X$ given by $f(t)=U_t$ is well-defined. From here it's not hard to show that this map is continuous and surjective, which completes the proof.
Questions
I'm trying to come up with a way (if it's even possible) to cover up a compact set by two closed balls, then cover up each of those closed with two more closed balls, etc in such a way that the diameters of these coverings decrease to zero. I suppose my conjecture is better stated more generally and formally is as follows:
Conjecture: Let $X$ be a nonempty compact metric space. There exists a family of compact sets $\big\{U_t\,:\,t\in\{0,1\}^{n},n\in\mathbb{N}\big\}$ such that $$ X=U_{0}\cup U_{1}, $$ and in general $$ U_t=U_{\{t,0\}}\cup U_{\{t,1\}} $$ for each $t\in\{0,1\}^n$. Moreover $$ \lim_{n\to\infty}\max_{t\in\{0,1\}^n}\text{diam}(U_t)=0. $$
Since $X$ is compact for each $n$ one could cover $X$ with finitely many closed balls $\mathscr{U}_n=\{U_{n,1},\ldots U_{n,k_n}\}$ of diameter say $2^{-n}$ so that $\mathscr{U}_{n+1}$ is a refinement of $\mathscr{U}_n$ for each $n$, but is there a way to alter this construction to where each refinement adds to more balls for each set?
This is part of a homework exercise so please only hints. Thanks in advance. Any help is greatly appreciated.
Hints: In the general case, I don't think it is wise to aim for precisely 2 balls in each refinement. Instead going from level n-1 to n you look at the max number, say $M$, of closed balls of radius $2^{-n}$ you need to cover a ball from level n-1. Then find $m$ so that $M\leq 2^m$. Now instead of using 1 binary symbol to describe the refinement you use m binary symbols. So your symbol sequence will consists of 'trunks' of binary symbols describing the nesting.
An example: Suppose you cover $X$ first by 3 balls, $U_0,U_1,U_2$. This may be encoded by 2 bits (as $2^2=4\geq 3$). So we map the 'cylinders' $[00],[01],[10]$ to $U_0$,$U_1$,$U_2$, respectively. The remaining $[11]$ you map to whatever you want, e.g. some arbitrary but fixed point $\xi\in X$. In the next level suppose you may cover with maximally 5 balls, so e.g. $U_1\subset U_{1,0},...,U_{1,4}$. You may encode all possibilities by 3 bits, so as before map e.g. $[01 \ 010]$ to $U_1\cap U_{1,2}$ (if non-empty). You create this way a map from $\{0,1\}^5=\{0,1\}^2\times \{0,1\}^3$ to the collection of intersection of balls at this level. Again for the leftover symbols (or if an intersection is empty) you map the cylinder to your chosen point $\xi$. Continuing this way, using compactness, you obtain the wanted continuous surjective map from $\{0,1\}^{\Bbb N}$ to $X$.
The above is not the most economical way of coding. A better, but for which more book-keeping is needed, is to let the number m depend on each ball at level n-1. But your binary tree at a given level $n$ will then not have uniform length in each branch.