Is this a topological closure operation? Does any relation 
$\propto\,\subseteq X\times \mathcal P(X)$ that extends $'\!\!\in'$ in the way that: 


*

*$x\in M\Rightarrow x\propto M$

*$\neg\exists x\in X:x\propto\emptyset$

*$x\propto A \subseteq B \Rightarrow x\propto B$

*$x\propto A\cup B\Rightarrow x\propto A \vee x\propto B$ 


defines a closure operation on subsets of $X$ by 
$x\in \overline M \Leftrightarrow x\propto M$?
My approach is: all the sets $\overline M$ satisfies the axioms for closed sets:


*

*$\emptyset$ is closed since $x\in\overline\emptyset\Leftrightarrow x\propto\emptyset
\Leftrightarrow x\in\emptyset$

*$X$ is closed since $x\in \overline X\Leftrightarrow x\propto X\Leftrightarrow x\in X$

*$x\in \overline{A\cup B}\Leftrightarrow x\propto A\cup B\Leftrightarrow x\propto A \vee x\propto B\Leftrightarrow x\in\overline A\cup\overline B\quad$ (3 & 4)

*$\displaystyle x\in\overline{\bigcap_iM_i}\Leftrightarrow x\propto\bigcap_iM_i\Leftrightarrow\forall i: x\propto M_i\Leftrightarrow\forall i: x\in \overline{M}_i\Leftrightarrow x\in\bigcap_i\overline{M}_i\quad$ (3)$\;\square$



I become unsure of the method when Hagen von Eitzen showed that a closure operation on $\mathcal P(X)$ I had defined with this method was faulty.
 A: Your fourth argument is incorrect. As Martin Sleziak mentioned in the comments, $x\propto M_i$ for each $i\in I$ does not imply that $x\propto\bigcap_iM_i$. Consider the relation $\propto$ defined on $\Bbb R\times\wp(\Bbb R)$ by $x\propto A$ iff $x\in\operatorname{cl}A$, where the closure is taken in the usual Euclidean topology; this relation evidently satisfies all of your axioms. However, if we set $M_0=(0,1)$ and $M_1=(1,2)$, then $1\propto M_0$ and $1\propto M_1$, but $1\not\propto\varnothing=M_0\cap M_1$.
Note that this doesn’t mean that $M\mapsto\overline M$ isn’t a closure operator: the ordinary closure operator in $\Bbb R$ doesn’t satisfy your fourth condition, either. The real question is whether your axioms imply that if $x\propto\overline M$, then $x\in\overline M$, so that $M\mapsto\overline M$ is idempotent. Unfortunately, they don’t. 
Define $\propto\subseteq\Bbb N\times\wp(\Bbb N)$ as follows: $n\propto A$ if and only if either 


*

*$n\in A$, or  

*$A\ne\varnothing$ and $n=\min(\Bbb N\setminus A)$.


It’s straightforward to check that $\propto$ satisfies conditions (1)-(4). However, $\overline{\{0\}}=\{0,1\}$, but $\overline{\{0,1\}\}}=\{0,1,2\}\ne\{0,1\}$: $2\propto\overline{\{0\}}$, but $2\notin\overline{\{0\}}$.
