Where $A$,$B$,$C$ are sets and $X$ is a set containing $A$ as a subset.
Both of these are from Proposition $3.1.27$ on Tao's Analysis I. These are the only proofs where I feel like I might've made a mistake in my logic because I'm rather inexperienced.
For the first I tried proving that every element $x$ in $A\cap (X\setminus A)$ is also an element of $\emptyset$ and vice versa. First suppose that $x\in \emptyset$ then the conclusion $x\in A\cap (X\setminus A)$ is vacuously true. Now suppose that $x \in A\cap (X\setminus A)$. Then $x \in A $ and $x\in X\setminus A \iff x \in A \text{ and } x\in \{y \in X: y\notin A\} \iff x\in A \text{ and } (x\in X \text{ and } x\notin A) $. Since we have $x\in A$ and $x \notin A$, the conclusion $x \in \emptyset$ is vacuously true.
For the second I tried proving it again by saying: We need to show that every element $x$ in $(A\cap B)\cap C$ is also an element of $A\cap (B\cap C)$ and vice versa; but I ended up proving an equivalance. Suppose $x \in (A\cap B)\cap C \iff (x\in A\cap b) \land (x\in C)\iff ((x\in A) \land (x\in B))\land (x \in C)\iff (x\in A)\land(x\in B)\land(x \in C) \iff (x\in A)\land((x\in B)\land(x\in C))\iff x\in A\cap(B\cap C)$.
So, both of these proofs kind of rely on the fact that $x\in A$ and $(x \in B \text{ and } x\in C)$ is the same as the statement $x\in A$ and $x \in B$ and $x\in C$. But this feels artificial. For the second proof I had the thought that set $\cap$ symbol has the same logic indentities as "and" so reducing the problem to logic made the solution trivial to the extent that it doesn't feel like a proof. I'm basically proving associativity of intersection by using the associativity of logic. But since the same logic applies to both symbols this seems like a non-proof.
To be clear, I'm not really interested in a new proof of these since I can do both of those in a more standard way that I'm pretty sure is correct. I want to know why these proofs do or don't work.
I think this is something I grappled with when working through Tao's analysis as well. This is the reasoning that I came down to accept in the end:
First a comment on the first proof: I don't fully understand your reasoning. The first inclusion is trivial as the emptyset is a subset of every set as you also point out. Suppose now $x \in A \cap (X \backslash A)$. Then, $x \in A$ and $x \in X \backslash A$. However, from here on I am not sure to follow your reasoning. Wouldn't it be easier to simply state that $x \in X \backslash A$ corresponds to that element which is in $X$ but not $A$; i.e., $x \in X \wedge x \notin A$. Thus we have $x \in A$ and $x \notin A$ which is already a contradiction? To be clear you might be stating this, but it wasn't clear to me how.
On to the second bit. I think the proof is corect and I do agree. There is a sense in which the associativity of $\cap$ boils down to the associativity of $\wedge$ and this might or might not feel artificial. However, this is essentially correct. Furthermore, you can prove this without ever writing down the logical symbol $\wedge$ as Tao does in some exercise, if I'm not wrong, and simply rely on words and the more intuitive meaning of words and sets. So that "something and something else" must be equivalent to "somthing else and something" by pure means of logic. This is also what, I think he tries to explain in the appendix A on mathemtical logic; i.e., to really understand these concepts rather than learn the rules; e.g., associativity of $\wedge$.
PS: Thinking about this more, I think that the issue comes in by mixing parts of the proof in plain English and part in first order logic. One should feel confortable to manipulate the ideas in both forms. In this case Tao seems to privilige the use of plain english and a more intuitive reliance on logic rather than a pure symbol manipulation. Obviosuly the two are equivalent, but the feeling of ariticiality might arise precisely because there is a mix of the two.