Below's my attempt at this proof. My question, in particular, is about line c. Does $\nvdash_\Sigma A$ imply $\vdash_\Sigma \lnot A$ for any wff $A$ in any normal modal system $\Sigma$? And if not, any hints on how I might correct the proof? Thanks so much.
We proceed contrapositively. So suppose (2) doesn't hold. We want to show that neither then does (1).
a. If (2) doesn't hold, then for some wff $A$, $\vDash A$, but $\nvdash A$.
b. Now, if $\nvdash A$, then $\vdash$ is consistent.
c. Also, if $\nvdash A$, then $\vdash\lnot A$.
d. But if $\vdash\lnot A$, then, by Soundness, $\vDash\lnot A$.
e. But if $\vDash A$, then $M,w\Vdash A$ for all worlds $w$ in all models $M$.
f. And similarly, if $\vDash\lnot A$, then $M,w\Vdash\lnot A$ for all worlds $w$ in all models $M$.
g. But by the definition of $M,w\Vdash$, $M,w\Vdash\lnot A$ iff $M,w\nVdash A$.
h. So there is no model $M$ such that for all worlds $w\in M$, $M,w\Vdash A$ and $M,w\Vdash\lnot A$.
i. So $A$ is unsatifisable.
j. So if (2) doesn't hold, (1) doesn't either.
Your concern is right: line c. "If $⊬A$, then $⊢¬A$" is not correct.
Consider an atom $p$: we have both $⊬ p$ and $⊬ ¬p$.
The proof must be: if $⊬A$, then $¬A$ is consistent (following your terminology) that means that $¬A$ is satisfiable (by hypotheses).
And this implies that $A$ is not valid, contradicting the assumption: $⊨A$.