Let $T$ be a set of formulas of a first-order language $L$. Show that $T$ is complete if and only if there is no sentence $A$ of $L$ such that both $T \cup \{ A \}$ and $T\cup \{\neg A\}$ are consistent.
I feel like the answer is staring me in the face and yet I can't seem to prove it.
Proof ($\Rightarrow$): Let $T$ be a complete theory. Hence for every sentence $\phi$, either $T \vdash \phi$ or $T \vdash \neg \phi$. Now let $\psi$ be any sentence. We show that either $T \cup \psi$ or $T \cup \neg\psi$ is inconsistent. By our assumption of $T$ being a complete theory, it follows that either $T \vdash \psi$ or $T \vdash \neg \psi$ must be the case. If the former, then $T \cup \neg\psi$ is inconsistent. If the latter, $T \cup \psi$ is inconsistent.
This shows only that if $T$ be a complete theory then either $T \cup \psi$ or $T \cup \neg\psi$ is inconsistent. Now, can you complete the proof showing that the other direction ($\Leftarrow$) also holds?