Something fishy in this algebra proof.

Question

I am trying to prove the existence of bases in the non finitely dimensional case. I want to apply Zorn's Lemma on the family of linearly independent sets.
Yet Ii feel like the logical construction of this proof has an error :
Let $\mathscr{A}$ be some chain, that has no maximal Element, then : $$\bigcup_{A \in \mathscr{A}} A \subsetneq B \;\text{ for some $B \in \mathscr{A}$ }$$ $B$ is linearly independant, therefore : $$\bigcup_{A \in \mathscr{A}} A \in \mathscr{A}$$
But since $B \in \mathscr{A}$, $B \subsetneq \bigcup_{A \in \mathscr{A}} A $ and therefore there is no such $B$. So there has to be a maximal element.
Now my problem is that from a logical standpoint, shouldn't it be that either our union isn't an element of the chain or there is a maximal element ?
So shouldn't I prove that our union is always an element of the chain for the proof to be complete ?

Answer 1

Zorn's Lemma only requires an upper bound for each chain and the upper bound need not belong to the chain.