BACKGROUND / MOTIVATION:
The other day, I was given a finite covering $\mathcal{B} = \{I_1, \ldots, I_n\}$ of $[0, 1]$ by open intervals, and I wanted to refine it by eliminating useless intervals, that is, those $I_k$ which are contained in the union of the others: $I_k \subseteq I_1 \cup \dots \cup \widehat{I_k} \cup \dots \cup I_n~~(\text{$I_k$ is ommited.)}$ Since my covering was finite, it was easy to establish an algorithm for doing this: check if $I_1$ is useless; if it is, remove it from the list and proceed to $I_2$, considering the new list now, and so on.
However, I couldn't help but notice that this process depends on the given order. For example, if $I_1 = [0, 1]$, $I_2 = [0, 3/4)$ and $I_3 = (1/4, 1]$, I would eliminate just $I_1$ and I'd be done. If, on the other hand, $I_3$ and $I_1$ switched places, my algorithm would end up eliminating $I_2$ and $I_3$ instead.
I then started wondering whether there is a more direct, non-algorithmic argument to show that such a refinement always exists. A naive guess would be to define $\mathcal{B}^{\prime} = \bigg\{ I_k \in \mathcal{B} : I_k \not\subseteq \bigcup_{j \neq k} I_j \bigg\},$ but this may be empty. In the finite case, one could settle for the given algorithm, but the situation becomes troublesome when dealing with arbitrary coverings. This has lead me to the following question.
QUESTION: Suppose you have a set $X$ and an arbitrary family of subsets $\mathcal{A} = \{A_i\}_{i \in I}$ of $X$ such that their union is $X$. How can you prove that there exists a refinement $\mathcal{A}^{\prime} = \{A_j\}_{j \in J}$, $J \subseteq I$, such that no member of $\mathcal{A}^{\prime}$ is useless (contained in some union of other members of $\mathcal{A}^{\prime}$) but $\mathcal{A}^{\prime}$ still covers $X$?
Thanks.