The successor of a set Definition : The successor of a set $x$ is the set $S(x) = x \bigcup \{x\}$ 
Prove that $x \subseteq S(x)$ and there is no $z$ such that $  x \subset z \subset S(x)$
I really battle with proofs :(
Here is what I have:
let $ y \in x $
then $y \in S(x)$
therefore $ x \subseteq S(x) $
I think I have over simplified something here and am not really proving anything :(
The second thing to prove is that there is no $z$ such that $ x \subset z \subset S(x) $
I really don't know how to approach this one.
Would it be by contradiction?
Assume that such $z$ exists, and then show that it can't exist?
I know that the statement $ x \subset z \subset S(x) $ means there is at least one element in $z$ that is not in $x$ and at least one element in $S(x)$ that is not in $z$ which implies that there are elements in $S(x)$ which are not in $x$.  Doesn't this contradict the definition of $S(x)$?
 A: You are correct about the first part, and your oversimplification is really a proof. For the second you are also correct when you proceed by contradiction. So there is an element in $z$ not in $x$ but in $S(x)$. That element can only be $\{x\}$. But then $z=S(x)$.
A: Your first step could be okay or not, depending on whether you understand why it’s justified. The point is that since $S(x)=x\cup\{x\}$, $x\subseteq S(x)$, simply because it’s always true that $a\subseteq a\cup b$, no matter what sets $a$ and $b$ are.
Now suppose that $x\subsetneqq z\subsetneqq S(x)$. Then there is some $u\in z\setminus x$. On the other hand, $z\subseteq S(x)$, so $u\in S(x)$. There is only one element of $S(x)$ that is not an element of $x$, so $z$ must be that element; what is it? Why does that contradict the assumption that $z\subsetneqq S(x)$?
A: Normally this construction is used to build the numbers "out of nothing", but for the purposes of clarity I will pretend that $x=\{a,b,c\}$.  Now, $S=x\cup \{x\}=\{1,2,3\}\cup\{\{1,2,3\}\}$, so $$S=\{a,b,c,\{a,b,c\}\}, \textrm{ a set with four elements}$$
To prove that $x\subseteq S$, you need to prove that each element of $x$ is an element of $S$ (or you can use that $A\subseteq A\cup B$ for any $B$).  Study the example above to convince yourself that $S$ differs from $x$ only by the inclusion of one extra element (possibly no extra elements if $\{x\}\in x$, but this is rare).  That should give you the intuition you need to complete the second part of the proof.
A: Straightforward proof of 2nd question
Suppose $x\subset z\subset S$ (1)
then $z=x\cup A$ where $A$ is a set such that $A \cap x =\emptyset$
Suppose $A \neq \emptyset$ for non triviality purposes
By definition of $S$ and using (1) $$ (x\cup A) \subset (x\cup \{x\})$$
Since $A \cap x =\emptyset$ , $A \subset \{x\}$
But $\{x\}$ is a single-element set
Hence $A=\{x\}$
Hence $z=S$
