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So what I came up with, similar to Cartesian product, is that for each element in S, you make an ordered pair with the whole set T. For each element in T, you make a pair with the whole set in S. Defined as follows:

$$S\mathbin{\square} T := \{(s,T) \mid \forall s \in S\} \cup \{(t,S) \mid \forall t \in T\}$$

where the ordered pair is defined as follows: $(x,t) = \{\{x\}, \{x,y\}\}$

To prove it is correct, I need to prove that the sets that are being unioned are disjoint, but not sure how to do that. Any help would be appreciated

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    It might not *be* disjoint, for example when $S = T$. – 2017-02-02
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    Here's a hint of a problem: what happens in your definition if you take S box S? – 2017-02-02
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    (Your ordered-pair idea is on the right track; but rather than using $S$ and $T$, which you don't know anything about, can you find two entities that you _know_ are different?) – 2017-02-02
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    Perhaps using the empty set and the set of natural/real numbers instead? – 2017-02-02
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    But then maybe S or T could be the empty set or the set of natural/real numbers – 2017-02-02
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    If you're feeling particularly rebellious, you might also consider $S\; \square\; T = \{1, 2, \ldots, |S| + |T|\}$ :) – 2017-02-02

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To disjointify them use a similar idea: define $0 = \emptyset$ as usual and $1= \{\emptyset\}$. These are clearly different.

Then $S\boxplus T := \cup \{(s,0): s \in S\} \cup \{(t,1): t \in T\}$ will do I think.

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    Which is the formal version of “paint the elements of $S$ in white and the elements of $T$ in black”. Choose your favorite colors, so long as they are distinct. – 2017-02-02
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    +1. Note that can even get a *commutative* operation with the desired property - that is, such that $S\Box T=T\Box S$! Let $S'=S-T$, $T'=T-S$, and $I=S\cap T$. We let $S\Box T$ be the (clearly disjoint) union of the following four sets: $$S'\times \{\{S\}\},\quad T'\times\{\{T\}\}, \quad I\times \{\{0, 1\}\}, \quad\mbox{ and }\quad I\times\{\{2, 3\}\}.$$ – 2017-02-02