As an exercise I'll try to reproduce a proof that I've seen, but my main question is about the act of relabeling sets which is used in the proof. The following is a proof that the exterior measure $m$ (star omitted for convenience) of a closed cube $Q \in \mathbb{R}^d$ is equal to its volume $|Q|$.
proof
Since $Q$ is a covering of itself we immediately have that $$mQ = \inf \sum_{j=1}^\infty |Q_j| \leq |Q| + |\emptyset| + \cdots = |Q|$$
So $mQ \leq |Q|$. Now we need to show the reverse inequality.
Let $Q$ be covered by an arbitrary covering $Q\subset \bigcup_{j=1}^{+\infty} Q_j$. For each $j$ choose an open cube $S_j$ such that
$$|S_j| \leq (1+\epsilon)|Q_j|\tag{1}$$
Clearly, $Q\subset \bigcup_{j=1}^{+\infty} S_j$ and since the union of the $S_j$ is an open set, we may choose from the collection of $S_j$ a finite subcovering to cover $Q$. After possibly relabeling* we may write $Q\subset \bigcup_{j=1}^{N} S_j$, so we have
\begin{align*} |Q| \leq \sum_{j=1}^N |S_j| &\leq \sum_{j=1}^N (1+\epsilon) |Q_j| \leq (1+\epsilon)\sum_{j=1}^{+\infty}|Q_j| \end{align*}
Since $\epsilon$ was arbitrary we have $|Q| \leq \sum_{j=1}^{+\infty}|Q_j|$And since we let the covering of $Q$ be arbitary we have
$$|Q| \leq \inf \sum_{j=1}^{+\infty}|Q_j|$$
which was required to be shown.
Regarding (1). Choose side length of
$$S_j = l_{S_{j}} = (1+\epsilon/2)^\frac{1}{d}l_{Q_j} \implies |S_j| = l_{S_j}^d = (1+ \epsilon/2)|Q_j| < (1+ \epsilon)|Q_j|$$.
*I don't see the reason to relabel. Can't we just assume, that a reasonable person would label a covering of closed cubes in a reasonable way. Is there an example of a way to number a covering of Q by closed cubes so that we could map as many counting numbers as we wanted to form the $S_j$ and never reach an $N$ where $Q$ is covered by $S$?