Proof: $A \neq \emptyset \implies A \nsubseteq \emptyset $ I proof that, let $A$ a set and $A \neq \emptyset$, then $A \nsubseteq \emptyset$; 
Proof by contradiction: if $A \subseteq \emptyset$ then by property I have an absurd , in fact by hypothesis $A \neq \emptyset$, therefore $A \nsubseteq \emptyset$.
Or, direct proof: by hypothesis $A \neq \emptyset$, then "$A \nsubseteq \emptyset$ or $ \emptyset \nsubseteq A$", but by property I have only case that $ A \nsubseteq \emptyset$...
Is it correct?
Thank you all in advance
 A: Or more positively, if $A\neq\emptyset$ then there exists some $x$ with $x\in A$, and since $x\notin\emptyset$ it follows that $A\not\subseteq\emptyset$.
A: There is a language issue so I can't quite tell if what you are suggesting is correct or not. In any case, it is not rigorously presented, so I provide you here with a rigorous proof. 
Suppose that $A\subseteq \emptyset $. Then, by definition of inclusion, for all $a\in A$ holds that $a\in \emptyset $. However, $a\in \emptyset$ never holds, and thus $a\in A$ never holds, so that $A=\emptyset $. By contrapositive it follows that if $A\ne \emptyset$ then $A$ is not contained in $\emptyset $.  
A: In Dijkstra-Feijen style, it is easy to rigorously prove that the two are even equivalent: for any set $A$,
$$
\begin{array}{ll}
& A ≠ ∅ \\
\equiv & \;\;\;\text{"expand $≠$ to introduce equality"} \\
& \lnot (A = ∅) \\
\equiv & \;\;\;\text{"introduce double inclusion -- suggested by shape of the desired conclusion"} \\
& \lnot (A ⊆ ∅ \land ∅ ⊆ A) \\
\equiv & \;\;\;\text{"$∅ ⊆ A$ is always true"} \\
& \lnot (A ⊆ ∅) \\
\equiv & \;\;\;\text{"simplify"} \\
& A ⊈ ∅ \\
\end{array}
$$
