I was trying to prove the following statement:
For $S \subseteq \mathbb{R}$, define $\mathbb{Q}(S)$ by:
$$ \mathbb{Q}(S) = \{q_0s_0 + \cdots + q_ks_k : k \in \omega, q_i \in \mathbb{Q}, \text{ and } s_i \in S \text{ for } i \leq k\} $$
$B \subseteq \mathbb{R}$ is a basis for $\mathbb{R}$ over $\mathbb{Q}$ if $B$ is $\subseteq$-minimal such that $\mathbb{R} = \mathbb{Q}(B)$, i.e. $\mathbb{R} = \mathbb{Q}(B)$ and for all $B' \subseteq B$ if $\mathbb{R} = \mathbb{Q}(B')$ then $B = B'$.
Now I can imagine you can prove this by just noting that $\mathbb{R}$ is a vector space over $\mathbb{Q}$ and thus has a basis in the linear algebra sense and showing this aligns with the basis being $\subseteq$-minimal. However, I came up with this alternate proof and was wondering about its validity.
Let $S_0 =\mathbb{R}$. It is evident that $\mathbb{Q}(S_0) = \mathbb{R}$, so if this is not a basis then there must be some $S_1$ s.t. $S_0 \supset S_1$ and $\mathbb{Q}(S_1) = \mathbb{R}$. Iterating this, either we will get a basis or we will have a chain $S_0 \supset S_1 \supset \ldots $ s.t. for any $i \in \mathbb{N}$, we have $\mathbb{Q}(S_i) = \mathbb{R}$. Then we can take the intersection $S_\omega = \bigcap\{S_i : i \in \mathbb{N} \}$ and $\mathbb{Q}(S_\omega) = \mathbb{R}$. Eventually we have to hit a basis otherwise we will exhaust all the elements in the set since we are indexing by ordinals.
This argument is exactly Zorn's lemma deconstructed for the $\supset$ relation and I was wondering if the reasoning is valid. More formally, you would consider the set $A = \{X : \mathbb{Q}(X) = \mathbb{R}\}$ and show that for any chain $B \subseteq A$, we have $\bigcap B \in A$ (i.e. $B$ has an upper bound) and thus the set $A$ must have a maximal element $S$, which is a basis for $\mathbb{R}$ over $\mathbb{Q}$.
This does not work. The problem is that the intersection of a chain in $A$ may not be an element of $A$. For a simple example, note that for each $x\in\mathbb{R}$, the set $X=(-\infty,x)$ is an element of $A$, and these sets form a chain. But the intersection of all of them is $\emptyset$, which is not in $A$.