How do I prove the anti-symmetry and there is a minimal element for each subset of $\alpha$ 
An ordinal number is a set $\alpha$ with the following properties:
(a) If $x,y \in \alpha,$ then either $x\in y$, $y\in x$, or $x=y.$
(b)If $y\in \alpha$ and $x\in y$, then $x\in \alpha$.

Let $\alpha$ be an ordinal number. For any two numbers $x$ and $y$ of $\alpha$, define $\leq$ on $\alpha$ by $x\leq y$ iff $x=y$ or $x\in y$.
Theorem:-
Let $\alpha$ is an ordinal number. Then $(\alpha,\leq)$ is a well ordered set.
Proof:-
I don't have the doubts in the proof of Reflexivity, Transitivity. I could be able to prove each element is comparable in $\alpha$.
Doubt 1. anti-symmetric
$x\leq y$ and $y\leq x$. Our claim is $x=y.$
Case 1 $x\leq y$ and $y\leq x$ means $x=y$ and $y\in x \implies x\in x$. Which is a paradox. How do I prove the anti-symmetry?
Case 2 $x\leq y$ and $y\leq x$ means $x=y$ and $y= x \implies x= y$.
Case 3 $x\leq y$ and $y\leq x$ means $x\in y$ and $y= x \implies x\in x$.
Case 4 $x\leq y$ and $y\leq x$ means $x\in y$ and $y\in x \implies x= y$.
Also How do I prove there is a minimal elemnt for each subset of $\alpha$?
 A: Anti-symmetry may be proven the following way: Let $x, y\in \alpha$ such that $x\leq y$ and $y\leq x$. This means that we have
$$
x\in y \text{ or } x = y\\
\text{and}\\
y\in x \text{ or }y = x
$$
So, one possibility is certainly that $x = y$. If $x\neq y$, then that means that we must have $x\in y$ and $y\in x$, which cannot be true (as the set $\{x, y\}$ would violate the axiom of foundation / regularity). This proves anti-symmetry.
As for well-orderedness, take a non-empty subset $S\subseteq \alpha$. By the axiom of foundation, there is an element $x\in S$ such that $x\cap S = \varnothing$. I claim that this $x$ is minimal in $S$.
Assume, for contradiction, that $x$ is not minimal. Then there is a $y\in S$ with $y\leq x, y\neq x$. This means that $y\in x$. But then we have $y\in x\cap S$, which is a contradiction.
A: I preassume that the axiom of regularity is accepted in this context.
Then consequently it cannot happen that $x\in x$ so the cases 1 and 3 will not show up.
Also the axiom excludes the situation that $x\in y\in x$ so also case 4 is excluded.
Our conclusion is then that we are dealing with case 2 as was to be shown.
Proved is now that $\leq$ is an anti-symmetric relation on $\alpha$.

Let $A$ be a non-empty subset of $\alpha$.
According to the axiom this set contains an element $a$ such that all elements of $a$ are not elements of $A$. 
So if $x\in A$ then we do not have: $x\in a$.
But what we do have is: $x\in a\vee x=a\vee a\in x$.
So what remains is: $x=a\vee a\in x$ or equivalently $a\leq  x$.
This proves that $a$ serves as least element of $A$ in the order $(\alpha,\leq)$.
