Let $S$ be a set in V (vector space). The following set is called dual of $S$: $$S^* = \{y \in V \mid \langle y,x\rangle \geq 0 \ \ \forall x \in S \}$$
Let $S_1,S_2$ be nonempty subsets of $V$.
I have proved that $S_1^* \cap S_2^* \subset (S_1+S_2)^*$, but unable to prove that when $0 \in S_1+S_2$ then $(S_1+S_2)^* \subset S_1^* \cap S_2^*$.
$$S_1+S_2 = \{x+y \mid \ x \in S_1, y \in S_2\}$$
As a counterexample, consider two subsets of $\mathbb R^n$, the unit ball $B$ and a set $U$ containing a single unit vector $u,\ \langle u,u\rangle=1$. The dual of the unit ball consists of only the zero vector, and the dual of $U$ is a half space. $$ B=\{x\in\mathbb R^n|\langle x,x\rangle\le 1\},\ \ \ B^*=\{0\} $$ $$ U=\{u\},\ \ \ U^*=\{x\in\mathbb R^n|\langle x,u\rangle \ge 0\}\supset\{0\} $$ The Minkowski sum of $B$ and $U$ is the unit ball shifted by $u$, which still contains the zero vector. By Cauchy-Schwarz, for vectors, $b\in B,u\in U$ the inner product $\langle u, u+b\rangle$ is nonnegative. This gives a nontrivial subset of the dual (this is in fact an equality, but there's no need to prove that here.): $$ (B+U)^*\supseteq\{\alpha u\ | \alpha\in\mathbb R,\ \alpha \ge 0\} $$ However, $B^*\cap U^*=\{0\}$. Thus the statement is false.
If we refine the hypothesis by requiring $0\in S_1\cap S_2$, the statement is true, and can probably be proven with more or less the same argument you found the first inclusion.