What's Wrong in this Proof Logic?

Question

Trying to show that the empty set $ \emptyset \subseteq A $, for any set $ A $.

Let $ x \in \emptyset $, then by definition, $ x \in \emptyset \iff (x \neq x) $.

$ x \in \emptyset \implies (x \neq x) \lor P $ where $ P $ is any statement

Let $ P $ be $ (x \in A) $, then $$ x \in \emptyset \implies (x \neq x) \lor (x \in A). $$ But $ (x \neq x) $ is false, then we can write $$ x \in \emptyset \implies (x \in A),$$ which is equivalent to $ \emptyset \subseteq A $.

Is this correct?

Answer 1

It is easier to note that $x \in \emptyset$ is false, hence $x \in \emptyset \Rightarrow x \in A$ is true for an arbitrary $x$. Every "$\Rightarrow$"-conclusion that you make based on a false premise is true trivially. This is easy to prove with the help of a truth table.

Answer 2

The "normal" proof is "vacuous" as they say.

If $X \not \subset A$ then there is some $x \in X$ and $x \not \in A$

So, is there any $x \in \emptyset$ and $x \not \in A$ ? No there isn't (because there is no $x \in \emptyset$ ), so $\emptyset \subset A$.