I'm new in convex analysis, and I'm working in the following Theorem.
Let $D_{i}\subset \mathbb{R}^{n}$, $i=1,2$ be a convex set such that $int(D_{i})\neq \emptyset$, $i=1,2$ and $int D_{1}\cap int D_{2} = \emptyset$. Then $D_{1}$ e $D_{2}$ are separated.
My attempt: Define $D= D_{1}-D_{2}$ and say that $int(D)= int(D_{1}) - int(D_{2})$, so $0\notin int(D)$, since $int D_{1}\cap int D_{2} = \emptyset$, there exists a hyperplane containing $\{0\}$ such that $int(D)$ is a subset of one of its corresponding open halfspaces. And, because $D\subset \overline{int(D)}$ , the closure of that halfspace contains $D$. Hence, there exists a (nonzero) $a\in \mathbb{R}^{n}$ such that $$ 0 \leq \inf\{ \langle a,x\rangle : x \in D \} = \inf\{ \langle a,x\rangle : x \in D_{1} \} − \sup \{ \langle a,x\rangle : x \in D_{2} \} $$, $$ 0 < \sup\{ \langle a,x\rangle : x \in D \} = \sup\{ \langle a,x\rangle : x \in D_{1} \} − \inf \{ \langle a,x\rangle : x \in D_{2} \}$$ this implies that $D_{1}$ and $D_{2}$ and be separated properly.
Actually I tried to modify other proposition that I find in the internet ,Here Theorem 4.3., but I don't know if it works. And I'm not sure for instance in properties like $int(D)= int(D_{1}) - int(D_{2})$ and $D\subset \overline{int(D)}$.
Thanks! in advance.