I am trying to prove that an arbitrary set $X$ is partial-ordered with respect to subset relation $\subset$ but I have some doubts. Indeed,
(i) $\forall x\in X$, $x\subset x$
(ii) $\forall x,y\in X$, if $x\subset y$ and $y\subset x$, then $x=y$
(iii) $\forall x,y,z\in X$, if $x\subset y$ and $y\subset z$, then $x\subset z$.
Is it correct to assume that $\forall x\in X$, $x$ is a subset? My answer is Yes only if $X$ is not empty by the foundation axiom. I appreciate any suggestion.
The empty set is a subset of itself. Indeed, the empty set is a subset of any set. This is proven as follows. Given a set $X$, suppose that $\varnothing\not\subset X$. This means that exists some object $x\in\varnothing$ such that $x\not\in X$, which is a contradiction, because the empty set has no elements. The existence of the empty set is assured by the empty set axiom.