Now, I was told that the above proof was valid by a professor. But I don't see how it could be valid as it is written. The only proof I can arguably see here is a proof that AUB[itex]\subseteq[/itex]BUA.

From the way its written, case 1 shows that A[itex]\subseteq[/itex]BUA while case 2 shows that B[itex]\subseteq[/itex]BUA; therefore, the conclusion would be AUB[itex]\subseteq[/itex]BUA.

You are right. But if we substitute A and B, then we also get a proof for the other inclusion. That is: a proof for the other inclusion follows from proving the first inclusion.

See I tired to point this out in class. The professor argued that my argument of
A→B and B→A therefore A=B was a totally different proof. And some how, he accomplishes the same thing without using this because of something to do with his description of an "arbitrary x".

First we prove (as in the OP) that [itex]E\cup F\subseteq F\cup E[/itex] for ALL sets E and F. This is what the OP does, right??

Now, we want to prove that [itex]A\cup B=B\cup A[/itex] for all sets A and B.
Well
[itex]\subseteq[/itex] follows if we take E=A and F=B.
[itex]\supseteq[/itex] follows if we take E=B and F=A.
So equality holds.

Formally, you indeed need to provide justification for both inclusions.

However, I wasn't present at the discussion, so I can't really say what your professor was trying to say. All I can say is that I think you have a good understanding of this situation and that what you say is correct.