If $C1\subset \mathbb{R}^n$ and $C_2\subset \mathbb{R}^n$ are two nonempty convex sets such that $C_1\cap C_2 \ne \emptyset$, can we prove the following result? :
If $\hat{x} \in \mathbb{R}^n$ is a vertex of $C_1$ such that $\hat{x} \in C_1 \cap C_2$, then $\hat{x}$ is also a vertex of $C_1 \cap C_2$.
My thoughts so far: $\hat{x} \in C_1 \cap C_2$ is a vertex of $C_1$ then it does not lie in any open line segment connecting two points of $C_1$. Then it certainly does not lie in any open line segment connecting two points of $C_1 \cap C_2$ . Hence it is a vertex of $C_1 \cap C_2$... but I am not certain how to show that x does not lie in any open line segment.. and how to present the proof formally.
If $x$ is a vertex of $C_1$, then there exists a hyperplane $H$ and a corresponding halfspace $H^+$ such that $H \cap C_1=\{\hat{x}\}$ and $C_1 \subset H^+$.
We claim that the same $H$ works.
Notice that $H \cap (C_1 \cap C_2) \subset H\cap C_1=\{\hat{x}\}$, furthermore, we are given that $\hat{x} \in C_1 \cap C_2$ and $\hat{x} \in H$, hence $\hat{x} \in H \cap (C_1 \cap C_2)$ and we can conclude that $H\cap (C_1 \cap C_2) = \{x\}$.
Clearly $C_1 \cap C_2 \subset C_1 \subset H^+$.
Hence $\hat{x}$ is a vertex of $C_1 \cap C_2$.