Q:

How to prove immediate predecessor by proof

My question says: 
Let A be a nonempty set and let P(A) be partially ordered by set inclusion.  Show that if B is in P (A) and x is not in B, then B is an immediate predecessor of B ∪ {x}. 
I started by saying:
let C be in A s.t.  B is a subset of C and C is a subset of B ∪ {x}.  
Suppose C does not equal B and C does not equal B ∪ {x}. 
Then C is a subset of B and B ∪ {x} is a subset of C. From the first subset since we assumed x is not in B then x is not in C. Now for the second subset since x is not in C, x is not in B ∪ {x}.  But this is a contradiction since x is in B ∪ {x}.  Then C=B or B ∪ {x}=C. 
update
let C be in A s.t.  B is a subset of C and C is a subset of B ∪ {x}.  
Suppose C does not equal B and C does not equal B ∪ {x}. 
Then We have proper sets B subset C and C subset B U {x}.  
Now let y be in C\B and z be in B U {x}\C. 
Then y is in C and y is not in B. z is in B U {x} and z is not in C
Then y,z is in (B U {x})\B. y in B or y in {x} and y not in B. z in B or z in {x} and z not in B.  So y,z in {x}.  y in {x} means y=x. z in {x} means z=x. Then y=x=z and y=z. This is a contradiction.  Thus B=C or B U {X} =C. B is a predecessor of B U {x}.  

A:

You have $B\subseteq C\subseteq B\cup\{x\}$, and you’re assuming (to get a contradiction) that $C\ne B$ and $C\ne B\cup\{x\}$. This is fine, but it does not imply that $B\cup\{x\}\subseteq C\subseteq B$, which is what your’e saying at the beginning of the last paragraph.  It says that $B\subsetneqq C\subsetneqq B\cup\{x\}$.
From this you can infer that there is some $y\in C\setminus B$ and some $z\in(B\cup\{x\})\setminus C$.  Then $y\ne z$ (why?),  and $y,z\in(B\cup\{x\})\setminus B$ (why?).  But then $y=x=z$, so $y=z$, which is the desired contradiction.