The question is about F6 of Section 8 ("Elementary facts about cones") in Donald Knuth's Sandwich Theorem (http://arxiv.org/pdf/math/9312214.pdf). He claims to prove $(A \cap B)^* = A^* + B^*$ when $A$ and $B$ are closed cones, but when I go through the proof, it seems like he only gets $$ A^* + B^* \subseteq (A \cap B)^* \subseteq (A^* + B^*)^{**}. $$ The last expression is equal to the closure of $A^* + B^*$.
I agree with the proof if and only if $A^* + B^*$ is closed. Why is $A^* + B^*$ closed?
I don't think you read the proof right. Not sure how to address your specific question so I'll just explain the proof instead.
The first step is to simply show that $(A \cap B)^* \supseteq A^* + B^*$ from first principles.
The rest of the proof is to show the reverse inclusion. He shows that $(A^* + B^*)^* \subseteq A^{**} + B^{**}$ by utilizing F2 and the fact that dual cones always contain the origin. Then he applies F2 again to obtain $(A^* + B^*)^{**} \subseteq (A^{**} \cap B^{**})^*$. But since both $A$ and $B$ are assumed to be closed cones, you can replace $A^{**}$ with $A$ and the same for $B$. You can also claim that $A^* + B^*$ is also a closed cone. Then you get $(A^* + B^*) \subseteq (A \cap B)^*$ thus completing the proof.