Prove that $a ⊈ {a}$, where $a$ is non-empty

Question

I show it like this.
Assume, $a subseteq {a}$
Let $x in a.~$ Then $x in {a}  (∵ a subseteq {a})$
And we know $a in {a}$
$therefore x in a in {a}$
$∴ x ∉ {a}$.
This is a contradiction!
$a subseteq {a}$ is false.$~∴ a ⊈ {a}$ is true.
Is this method right or wrong? If there is another way to prove it?

egreg · Answer

No, it is incorrect. For one thing: the inclusion $asubseteq{a}$ is obviously true if $a=emptyset$.
If you assume that $aneemptyset$, then your idea works, but for a different reason than you believe. If $xin a$, then $xin{a}$, by transitivity of inclusion, hence $x=a$. Thus $ain a$, which is ruled out by the axiom of foundation (or regularity).
If the axiom of foundation is not assumed, then it is impossible to prove the statement (when restricted to nonempty $a$), because there might exist sets such that $a={a}$.

Prove that $a ⊈ {a}$, where $a$ is non-empty

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