Proving $A \cup B \cup C = (A-B)\cup(B-C)\cup(C-A)\cup(A \cap B \cap C)$ - not sure why this reasoning is failing. This seems like a fairly elementary problem but I am not sure where my reasoning is incorrect, whether it's down to a simple error in Boolean algebra or if I'm misunderstanding something else.
We want to show that

$A \cup B \cup C = (A-B)\cup(B-C)\cup(C-A)\cup(A \cap B \cap C)$.

Let $x \in A \cup B \cup C$. Then:
$(x \in A) \lor (x \in B) \lor (x \in C) \\
\leftrightarrow (x \in A \land \mathbf T) \lor (x \in B  \land \mathbf T) \lor (x \in C  \land \mathbf T) \\
\leftrightarrow (x \in A \land (x \in B \lor x\notin B)) \lor (x \in B  \land (x \in C \lor x\notin C)) \lor (x \in C  \land (x \in A \lor x\notin A)) \\
\leftrightarrow ((x \in A \land x \in B) \lor (x\in A \land x\notin B)) \lor ((x \in B \land x \in C) \lor (x\in B \land x\notin C)) \lor ((x \in C \land x \in A) \lor (x\in C \land x\notin A)) \\
\leftrightarrow (x \in A \land x \in B) \lor (x\in(A-B)) \lor (x \in B \land x \in C) \lor (x\in (B-C)) \lor (x \in C \land x \in A) \lor (x\in (C- A)) \\
\leftrightarrow x\in(A-B) \lor x\in (B-C) \lor x\in (C- A) \lor (x \in A \land x \in B) \lor(x \in B \land x \in C) \lor (x \in C \land x \in A)$
...which almost works, but obviously $(x \in A \land x \in B) \lor(x \in B \land x \in C) \lor (x \in C \land x \in A) \neq x\in A \land x\in B \land x\in C$.
I suspect I am erroneously equivocating some Boolean algebra operation onto set operations.
 A: From your third line, continue:
$((x \in A \land x \in B) \lor (x\in A \land x\notin B)) \lor ((x \in B \land x \in C) \lor (x\in B \land x\notin C)) \lor ((x \in C \land x \in A) \lor (x\in C \land x\notin A)) \Leftrightarrow \text{ (removing redundant parentheses)}$
$(x \in A \land x \in B) \lor (x\in A \land x\notin B) \lor (x \in B \land x \in C) \lor (x\in B \land x\notin C) \lor (x \in C \land x \in A) \lor (x\in C \land x\notin A) \Leftrightarrow \text{ same trick you did at start!}$
$(x \in A \land x \in B \land \mathbf T) \lor (x\in A \land x\notin B) \land \mathbf T) \lor (x \in B \land x \in C \land \mathbf T) \lor (x\in B \land x\notin C \land \mathbf T) \lor (x \in C \land x \in A \land \mathbf T) \lor (x\in C \land x\notin A \land \mathbf T) \Leftrightarrow$
$(x \in A \land x \in B \land (x \in C \lor x \notin C)) \lor (x\in A \land x\notin B) \land (x \in C \lor x \notin C)) \lor (x \in B \land x \in C \land (x \in A \lor x \notin A)) \lor (x\in B \land x\notin C \land (x \in A \lor x \notin A)) \lor (x \in C \land x \in A \land (x \in B \lor x \notin B)) \lor (x\in C \land x\notin A \land (x \in B \lor x \notin B)) \Leftrightarrow$
$(x \in A \land x \in B \land x \in C) \lor (x \in A \land x \in B \land x \notin C) \lor (x\in A \land x\notin B \land x \in C) \lor (x\in A \land x\notin B \land x \notin C) \lor (x \in B \land x \in C \land x \in A) \lor (x \in B \land x \in C \land x \notin A) \lor (x\in B \land x\notin C \land x \in A) \lor (x\in B \land x\notin C \land  x \notin A) \lor (x \in C \land x \in A \land x \in B) \lor (x \in C \land x \in A \land x \notin B) \lor (x\in C \land x\notin A \land x \in B) \lor (x\in C \land x\notin A \land  x \notin B) \Leftrightarrow \text{ Commutation }$
$(x\in A \land x\notin B \land x \in C) \lor (x \in C \land x \in A \land x \notin B) \lor (x\in A \land x\notin B \land x \notin C)  \lor (x \in A \land x \in B \land x \notin C) \lor (x\in B \land x\notin C \land x \in A) \lor (x\in B \land x\notin C \land  x \notin A) \lor (x\in C \land x\notin A \land x \in B) \lor (x \in B \land x \in C \land x \notin A) \lor (x\in C \land x\notin A \land  x \notin B) \lor (x \in A \land x \in B \land x \in C) \lor (x \in B \land x \in C \land x \in A) \lor (x \in C \land x \in A \land x \in B) \Leftrightarrow \text{ Removing duplicates by Idempotence }$
$(x\in A \land x\notin B \land x \in C) \lor (x\in A \land x\notin B \land x \notin C) \lor (x\in B \land x\notin C \land x \in A) \lor (x\in B \land x\notin C \land  x \notin A) \lor (x\in C \land x\notin A \land x \in B) \lor (x\in C \land x\notin A \land  x \notin B) \lor (x \in A \land x \in B \land x \in C) \Leftrightarrow $
$(x\in A \land x\notin B \land (x \in C \lor x \notin C)) \lor (x\in B \land x\notin C \land (x \in A \lor x \notin A)) \lor (x\in C \land x\notin A \land (x \in B \lor x \notin B)) \lor (x \in A \land x \in B \land x \in C) \Leftrightarrow $
$(x\in A \land x\notin B \land \mathbf T) \lor (x\in B \land x\notin C \land \mathbf T) \lor (x\in C \land x\notin A \land \mathbf T) \lor (x \in A \land x \in B \land x \in C) \Leftrightarrow $
$(x\in A \land x\notin B ) \lor (x\in B \land x\notin C ) \lor (x\in C \land x\notin A ) \lor (x \in A \land x \in B \land x \in C) \Leftrightarrow $
$ (x\in B - C) \lor (x\in A - B) \lor (x\in A - C) \lor (x \in A \land x \in B \land x \in C)$
Now, that's pretty lengthy ... can we shorten this?
Yes, first of all, instead of doing something like:
$x \in A \Leftrightarrow x \in A \land T \Leftrightarrow x \in A \land (x \in B \lor x \notin B) \Leftrightarrow (x \in A \land x \in B) \lor (x \in A \land x \notin B)$
there is a handy-dandy equivalence called:
Adjacency
$P \Leftrightarrow (P \land Q) \lor (P \land \neg Q)$
and with that, you can just do this in 1 step:
$x \in A \overset{Adjacency}\Leftrightarrow (x \in A \land x \in B) \lor (x \in A \land x \notin B)$
Moreover, all boolean logical rules have a perfect equivalent in set-theory, so that, for example, you also have:
$A \overset{Adjacency}= (A \cap B) \cup (A \cap B^C)$
So, while still following the same basic thought, we can do your problem as follows:
$A \cup B \cup C=$
$(A \cap B) \cup (A \cap B^C) \cup (B \cap C) \cup (B \cap C^C) \cup (C \cap A) \cup (C \cap A^C)=$
$(A \cap B \cap C) \cup (A \cap B \cap C^C) \cup (A \cap B^C \cap C) \cup (A \cap B^C \cap C^C) \cup (B \cap C \cap A) \cup (B \cap C \cap A^C) \cup (B \cap C^C \cap A) \cup (B \cap C^C \cap A^C) \cup (C \cap A \cap B) \cup (C \cap A \cap B^C) \cup (C \cap A^C \cap B) \cup (C \cap A^C \cap B^C)=$
$(A \cap B^C \cap C)  \cup (C \cap A \cap B^C) \cup (A \cap B^C \cap C^C) \cup (A \cap B \cap C^C) \cup (B \cap C^C \cap A) \cup (B \cap C^C \cap A^C) \cup (C \cap A^C \cap B) \cup (B \cap C \cap A^C) \cup (C \cap A^C \cap B^C) \cup (A \cap B \cap C) \cup (B \cap C \cap A) \cup (C \cap A \cap B) =$
$(A \cap B^C \cap C)  \cup (A \cap B^C \cap C^C) \cup (B \cap C^C \cap A) \cup (B \cap C^C \cap A^C) \cup (C \cap A^C \cap B) \cup (C \cap A^C \cap B^C) \cup (A \cap B \cap C) =$
$(A\cap B^C)\cup(B\cap C^C)\cup(C\cap A^C)\cup(A \cap B \cap C)=$
$(A-B)\cup(B-C)\cup(C-A)\cup(A \cap B \cap C)$
But, we can do better yet. Consider:
Reduction
$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$
which is a special case of:
Generalized Reduction
$(P \land Q) \lor (P \land \neg Q \land R) \Leftrightarrow (P \land Q) \lor (P \land R)$
With these, your identity is more quickly shown, and probably better best understood when starting with the right side:
$(A-B)\cup(B-C)\cup(C-A)\cup(A \cap B \cap C)=$
$(A\cap B^C)\cup(B\cap C^C)\cup(C\cap A^C)\cup(A \cap B \cap C)\overset{Generalized Reduction}=$
$(A\cap B^C)\cup(B\cap C^C)\cup(C\cap A^C)\cup(A \cap C)\overset{Adjacency}=$
$(A\cap B^C)\cup(B\cap C^C)\cup C\overset{Reduction}=$
$(A\cap B^C)\cup B\cup C\overset{Reduction}=$
$A \cup B\cup C$
A: In showing that $[A \cup B \cup C] \subseteq [(A-B)\cup(B-C)\cup(C-A)\cup(A \cap B \cap C)],$ I would use a case system:
Case 1: $x\in(A-(B\cup C))$. Then show that $x\in A-B$. 
Case 2: $x\in ((A\cap B)-C)$. Then show that $x\in B-C$. 
Case 3: $x\in A\cap B\cap C$. This one's straight-forward.
Once you've done these three cases, then WLOG, you've shown the one direction to be true. 
In showing that $[A \cup B \cup C] \supseteq [(A-B)\cup(B-C)\cup(C-A)\cup(A \cap B \cap C)],$ again, I would take cases. Just take each disjunct separately:
Case 1: $x\in A-B$. Then show it's in $A\cup B\cup C$. $B-C$ and $C-A$ will be similar.
Case 2: $x\in A\cap B\cap C$. Then it'll be straight-forward to show that $x\in A\cup B\cup C$. 
A: Shesh.   Using inclusion arguments inline unneccessarily bloats the text; a source for error and confusion.
Your proof essentially boils down to:
$\def\stm{{\smallsetminus}}\begin{align}\quad & A\cup B\cup C
\\ \tag 1 =~&(A \stm B)\cup(A\cap B)\,\cup\,(B\stm C)\cup(B\cap C)\,\cup\,(C\stm A)\cap (A\cap C)
\\ \tag 2 =~&(A\stm B)\cup(B\stm C)\cup(C\stm A)\,\cup\,(A\cap B)\,\cup\,(B\cap C)\,\cup\,(A\cap C)
\\~~~ &\text{...and from there move on to:}
\\ \tag 3 =~&(A\stm B)\cup(B\stm C)\cup(C\stm A)\,\cup\,((A\cap B)\stm C)\cup((B\cap C)\stm A)\cup((A\cap C)\stm B)\,\cup\, (A\cap B\cap C)
\\ \tag 4 =~&(A\stm B)\cup((A\cap C)\stm B)\,\cup\,(B\stm C)\cup((A\cap B)\stm C)\,\cup\,(C\stm A)\cup((B\cap C)\stm A)\,\cup\, (A\cap B\cap C)
\\ \tag 5 =~&(A\stm B)\cup(B\stm C)\cup(C\stm A)\,\cup\, (A\cap B\cap C)
\end{align}$
If you feel the need, you may use inclusion arguments to seperately prove the justifications for steps in (1), (3), and (5). 
$${{(1): X=(X\stm Y)\cup(X\cap Y)\\ (2): \text{Commutation}\\ (3): (X\cap Y)=((X\cap Y)\stm Z)\cup(X\cap Y\cap Z)\text{ plus distributing the common disjunct}\\ (4) :\text{commutation and association}\\(5): (X\stm Z)\cup((X\cap Y)\stm Z)=(X\stm Z)}}$$

PS: Note that indeed clearly $(A\cap B)\cup(B\cap C)\cup(A\cap C)\neq A\cap B\cap C$ in general, but ... when $X\cup Y\neq Y$ you can still have $(Z\cup X)\cup (X\cup Y) =(Z\cup X)\cup Y$ .
A: You proved
$A \cup B \cup C$ = 
$(A - B) \cup (B - C) \cup (C- A)$ $\cup$
$(A \cap B) \cup (B \cap C) \cup (C \cap A)$
you can make a Venn-diagram to follow this:
Let $X = A \cap B \cap C$
Then 
$(A\cap B) \cup (B \cap C) \cup (C \cap A)$ = 
$[((A\cap B) - X) \cup ((B \cap C) - X) \cup ((C \cap A) - X)] \cup X$ ........ (1)
We have:
$((A\cap B) - X) \subset (B - C)$ ........ (2a)
$((B\cap C) - X) \subset (C - A)$ ........ (2b)
$((C\cap A) - X) \subset (A - B)$ ........ (2c)
Therefore:
$(A - B) \cup (B - C) \cup (C - A)$ $\cup$ 
$(A\cap B) \cup (B \cap C) \cup (C \cap A)$ = (by (1)) =
$(A - B) \cup (B - C) \cup (C - A)$ $\cup$
$[((A\cap B) - X) \cup ((B \cap C) - X) \cup ((C \cap A) - X)] \cup X$ =
$(A - B) \cup (B - C) \cup (C - A)$ $\cup$
$[((B\cap C) - X) \cup ((C \cap A) - X) \cup ((A \cap B) - X)] \cup X$ = (by 2a,b,c) =
$(A - B) \cup (B - C) \cup (C - A) \cup X$ =
$(A - B) \cup (B - C) \cup (C - A) \cup (A \cap B \cap C)$
Conclusion:
$A \cup B \cup C$ = 
$(A - B) \cup (B - C) \cup (C- A)$ $\cup$ $(A \cap B \cap C)$
