Existence of a Minimal Cover

Solution 1:

If $X$ is an uncountable null set then it cannot have a minimal cover. A fat Cantor set cannot have a minimal cover either since every non-trivial closed cube will leak out of it.

If $X \subseteq \mathbb{R}^d$ has finite outer measure and $\{Q^{\star}_n : n \geq 1\}$ is a minimal cover of $X$, then $X$ is a full outer measure subset of $Y$ where $Y = \bigcup \{Q^{\star}_n : n \geq 1\}$. Isn't this a characterization of such sets - namely sets with full outer measure in a countable union of cubes? Or sets which are a union of countably many full outer measure subsets of cubes?

On real line this just amounts to being a full outer measure subset of a union of an open set and a countable set.

Solution 2:

In particular, there is a minimal cover for any open set. This follows since any open set can be partitioned into a countable union of half-open cubes of the form $\bar{x}/2^i + [0,1/2^i)^k$ where $\bar{x} \in \mathbb{Z}^k$. The closures of these cubes would give the minimal covering.