In the reference provided by Mike, the proof of the equivalence between $\mathrm{Ker}(S)$ and the intersection of the inclusion-maximal convex subsets of $S$ seems to rely on the following simple argument: "Since, for all $x \in S$, $\{x\}$ is convex, there must be some inclusion-maximal convex subset $X$ of $S$ such that $x\in X$." I believe that any set $S$ (in a real vector space) is indeed covered by its inclusion-maximal convex subsets, but I cannot see that such a simple argument holds as a proof.
Try to replace convexity by some other set property, e.g. countability, and define inclusion-maximal subsets w.r.t. this property as well. By applying the simple argument, we get, since $\{x\}$ is countable, that every $x$ is in some inclusion-maximal countable subset of $S$. This is obviously wrong, since unless $S$ is itself countable, $S$ does not have any inclusion-maximal countable subsets (if $X$ is countable then also $X\cup \{y\}$ for any $y\in S\setminus X$ is).
I suspect that proving coverage by maximal subsets relies more heavily on convexity. The property that singletons are convex seems insufficient.