How to prove Baire Category theorem by Zorn's lemma? Baire Category theory states that, in a complete metric space, the union of countably many dense open sets is dense. The proof relies on Axiom of choice. But AC is equivalent to Zorn's Lemma.
So can anyone produce a proof, using Zorn's Lemma instead? 
 A: If you can prove something using choice, then of course you can prove it using Zorn's lemma. Simply prove choice first, then repeat the proof. You can even fold the proofs together and get a slightly more direct proof.
Here is a convoluted outline of such proof. We have our dense open sets, $U_n$, we may assume they are decreasing in inclusion, of course. To show the intersection is dense, pick a non-empty open set $W$ and define a sequence of points $x_n\in U_n$, such that the sequence converges to a point in $W$.
To use Zorn's lemma here, consider the partial order of possible finite sequences for this proof. If every such sequence is finite, then every chain is bounded from above (trivially) and there is a maximal element. But an easy argument shows that no maximal element can exist, because of density and openness of the $U_n$'s. Therefore there is a chain without an upper bound, which must mean it is infinite. And this provides the sequence we want.

The lesson to learn here is that choice has many variants and equivalents and weak forms. And using one kind of equivalent might be simpler in some proofs than another.
