If $\Gamma$ is a maximal K-consistent set of formulas, where K is the minimal modal logic, is $\Gamma$ closed under the necessitation rule? That is, if $\phi\in\Gamma$, do I necessarily have $\Box\phi\in\Gamma$?
I am asking because in the book I am reading--Modal Logic for Open Minds by Van Bentham--the claim is made that such sets are closed under K-derivable formulas, which, under my reading, implies closure under the necessitation rule. This is not necessary to prove the completeness of K, which is the whole reason consistent sets were introduced in the first place, but I would still like to know whether this holds or not.
I think there is a small mix-up in your question. If some formula $\varphi \in \Gamma$, then it is not necessarily the case that $\square \varphi \in \Gamma$. The reason for this is that the necessitation rule does not preserve truth (while it preserves validity), i.e. if $\varphi$ is some true formula, and not a validity, then we cannot infer $\square \varphi$. A simple example: $M_s \models p$ does not imply $M_s \models \square p$.
At the same time, what is meant in the textbook you mention is that theorems of $\mathrm{K}$ are closed under necessitation, i.e. all formulas that are derivable from the axioms of $\mathrm{K}$ via application of rules of inference. In this case, set $\mathbb{K}$ of all theorems derivable from the axiom system $\mathrm{K}$ belongs to the MCS $\Gamma$.