Prove : Let A be a set. Then there is a set B such that B /∈ A.

Prove:

Let A be a set. Then there is a set B such that B /∈ A.

Proof Consider the statement ‘x is a set and x /∈ x.’ By the axiom of specification, there is a unique set B ={x ∈ A such that x is a set and x /∈ x}. We show that B /∈ A. Suppose that B ∈ A.If B ∈ B, then B /∈ B. Next, if B /∈ B, then since B ∈ A (supposition), and B is a set, B ∈ B.Thus,‘B /∈ B if and only if B ∈ B.’ This is a contradiction (P if and only if--P is a contradiction) to the supposition that B ∈ A. Hence, B /∈ A.

Here the statement taken is already a contradiction which leads to B/∈A . Can't we do this in any other manner?

You don't say what axioms you have to hand, but let's think about this anyway. The statement fails if $A$ is the set of all sets, assuming we're in a setting where such a thing exists. (For example, in Quine's "New Foundations", we do have a set of all sets, which contains itself as well as all the other sets.) But if you have the axiom of regularity, $$\forall x: x \neq \emptyset \implies \exists y:(y \in x \wedge y\cap x = \emptyset),$$ then take $x = \{A\}$ to find that there exists some element $y$ of $\{A\}$ such that $y \cap \{A\}$ is empty. This $y$ can only be $A$ itself. So $A \cap \{A\}$ is empty, from which it follows that $A \not\in A.$

(For this argument, we do also need to be able to show that $\{A\}$ is a set! If we have the axiom of pairing then we could apply that to $A$ and $A$.)

In the context of ZF o GBN, that's a neat idea, and I like it because it only uses specification. Another one would be using the fact that axiom of foundation + axiom of pairing prove $\forall A,A\notin A$; for if $A\in A$, then $\{A\}$ intersects all its elements.