No. $S$ is always convex by assumption. You only drop the non-active constraints in the proof, exactly at those points, where the derivatives $g_i'(x)$ appear.

Then, you write down the same proof, but leave out the non-active constraints. This is typically done by introducing the set $A(x) = \{i : g_i(x) = 0\}$ of active constraints.

Since you have strict inequalities in your $S$, the KKT conditions are $\nabla f(\bar x) = 0$ and by convexity of $f$, this is a global minimizer. Did you mean non-strict inequalities? Then, the result should hold, maybe even without a CQ.

What does $BX \le c$ mean with matrices $B$, $X$ and a vector $c$?

@Denoising: How about $\bar u = \bar u_n \equiv 0$? This satisfies 1-3.