= Solution
Order the consistent sets of sentences extending $\Sigma$ by inclusion. This collection is nonempty. The union of a chain is consistent: any derivation of a contradiction uses finitely many assumptions, which all lie in one member of the chain. <Zorn's lemma> therefore supplies a maximal member, which is a maximal consistent extension of $\Sigma$.
For a <countable> language one can instead enumerate all sentences and successively add either a sentence or its <negation> while preserving consistency. At least one choice is consistent, since inconsistency of both extensions, together with the <deduction theorem for propositional logic> and classical reasoning, would make the previous stage inconsistent. The union of the stages is consistent by finite use of assumptions and decides every sentence, so it is maximal consistent.
The chain proof covers arbitrary set-sized languages; no restriction on the language <cardinality> was printed here. The later prohibition on choice applies to the cardinal-arithmetic question, not this logic construction.
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