I have to proof by contradiction that $ A \subseteq A $; if $ A \nsubseteq A $ then $ \exists x \in A ( x \notin A ) $ but $ \exists x \in A ( x \notin A ) $ is contradiction (in fact: $ \exists x ( x \in A \wedge x \notin A ) $) therefore $ A \subseteq A $ is true... The process is correct? Thank you all in advance
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Your proof is correct, the way I would write it is as follows:
Suppose there exists a set $A$ such that $A \nsubseteq A$. Then there exists $x \in A$ such that $x \notin A$. But $x \in A$, contradiction. Hence $A \subseteq A$.
The extra formalism isn't necessary unless you are trying to get a better understanding of using the quantifiers, which in this case you have used correctly.
It looks fine but it is better you clarify both $A$ as $$A_1\subseteq A_2$$ where $A_1=A_2=A$.