We place ourselves in $\Bbb{R}^d$, for $d\geq 1$. Let $\mathcal{h}$ a hyperplane, and let $S$ be a finite set of points that all lie in one of the (closed) half-spaces limited by $h$.
Let $C$ denote transformation that to one set assigns its convex hull.
Prove that $C(S)\cap h=C(S\cap h )$.
As $S \cap h \subseteq h$ we have $C(S \cap h) \subseteq C(h)$ as $h$ is convex. And obviously $C(S \cap h) \subseteq C(S)$. Therefore $C(S \cap h) \subseteq C(S) \cap h$.
Regarding the converse inclusion. As $S$ is included in a half space having $h$ for boundary, a point $x$ belongs to $C(S) \cap h$ only if is is a convex combination of points of $S$. Otherwise it’s coordinate over a line not included in $h$ won’t vanishes in contradiction with $x \in h$. Hence $x$ is a convex combination of points of $S \cap h$ which means that $x \in C(S \cap h)$.