Denote the boxes as $a_{1},a_{2},a_{3},a_{4},b_{1},b_{2},b_{3},b_{4},c_{1},c_{2},c_{3},c_{4}$. Let $A=\lbrace{a_{1},a_{2},a_{3},a_{4}\rbrace}, B=\lbrace{b_{1},b_{2},b_{3},b_{4}\rbrace}, C=\lbrace{c_{1},c_{2},c_{3},c_{4}\rbrace}$. Let $X$ be the set of boxes that the audience opens. Now the assistant opens the box according to the following rule. $\bullet$ Assistant opens $c_{1}$ if $X\in \lbrace{a_{1},a_{2},b_{1},b_{3}\rbrace}$. $\bullet$ Assistant opens $c_{2}$ if $X\in \lbrace{a_{1},a_{2},b_{2},b_{4}\rbrace}$. $\bullet$ Assistant opens $c_{3}$ if $X\in \lbrace{a_{3},a_{4},b_{1},b_{4}\rbrace}$. $\bullet$ Assistant opens $c_{4}$ if $X\in \lbrace{a_{3},a_{4},b_{2},b_{3}\rbrace}$. And the same rules apply symmetrically to $X\in B\cup C$ or $X\in C\cup A$. Then we can see that the magician's trick can always work.

Thanks!Had some tough time trying to solve this but your answer is really helpful!

Variant with 13 boxes: Quote: A magician and his assistent are performing the following trick.There is a row of 13 empty closed boxes. The magician leaves the room, and a person from the audience hides a coin in each of two boxes of his choice, so that the assistent knows which boxes contain coins. The magician returns, and the assistant is allowed to open one box that does not contain a coin. Next, the magician selects 4 boxes, which are simultaneously opened. The goal of the magician is to open both boxes that contain coins. Does there exist a method that will allow the magician and his assistant to always successfully perform the trick?