Let $X$ and $Y$ be two sets.
$X$ is a subset of $Y$, written $X \subseteq Y$, if
\[\forall x :\enspace x \in X \enspace\rimp\enspace x \in Y.\]$X$ is a proper subset of $Y$, written $X \subset Y$, if $X \subseteq Y$ and $X \neq Y$.
$\empt$ is the only set that is a subset of every non-empty set.
Proof. $\empt$ is a subset of every set since no $x \in \empt$. If $X$ is a subset of every non-empty set, then $X \subseteq \{X\}$. By irreflexivity of $\in$, $X \neq \{X\}$, so $X = \empt$.
For any set $X$,
\[X \subseteq X.\]
For any sets $X$, $Y$ and $Z$,
\[X \subseteq Y \,\land\, Y\subseteq Z \enspace\rimp\enspace X \subseteq Z.\]
For any sets $X$ and $Y$,
\[X \subseteq Y \,\land\, Y \subseteq X \enspace\rimp\enspace X = Y.\]
As a corollary of reflexivity, transitivity and antisymmetry of $\subseteq$:
For any set $X$, $(\powerset(X), \subseteq)$ is a partial order.