Proving the lemma from the axiom of choice is an application of Zorn's Lemma, and is fairly standard as these things go. The partial ordering is simply that of a subset. The non-trivial part is proving that a maximal filter contains every set or its complement. Let us say F contains neither A nor X \\ A. From maximality, that means there is a set B in F such that the intersection of A and B is empty (otherwise, the union of F and {A} would generate a filter). Likewise, there is a C such that the intersection of C and X \\ A is empty. The intersection of C and B (let us call it D) is in F. D has empty intersection with both A and X \\ A, so it has an empty intersection with X, so it is empty. But a filter cannot contain an empty set.
This proof uses Zorn's Lemma, which is equivalent to the axiom of choice. The Ultrafilter Lemma cannot be proven from ZF (the Zermelo-Fraenkel axioms) alone, and it cannot be used to prove the axiom of choice, so it is properly weaker.