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
$x\in \overline M \Leftrightarrow x\propto M$, since all the sets $\overline M$ satisfies the axioms for closed sets.

Given a set $X$. For $M\subseteq X$ and $\mathcal M \subseteq\mathcal P(X)-\{\emptyset\}$ define:

$(1)\quad M\propto\mathcal M\Leftrightarrow M\cap\displaystyle\bigcup_{L\in\mathcal M}L\ne\emptyset$

that satisfies 1-4 above and therefor define a topology on $\mathcal P(X)-\{\emptyset\}$.

Given a topological space $(X,\tau)$, define:

$(2)\quad M\propto\mathcal M\Leftrightarrow \overline M\cap\displaystyle\bigcup_{L\in\mathcal M}\overline L\ne\emptyset$

that also satisfies 1-4 and therefore define a topology on $\mathcal P(X)-\{\emptyset\}$. If $\tau$ is the discrete topology, then $(1)$ coincide with $(2)$. In this construction the empty set must be an isolated point.

**Old answer**:

Given a topological space $\langle X,\tau\rangle$. For $\alpha\in 2^{2^X}\!$ and $M\in 2^X\!$, define

(1) $\quad$ $M\in\overline{\alpha} \Leftrightarrow
\exists L\in\alpha: \overline{L}\cap \overline{M}\ne\emptyset$.

This closure define a topological space $\langle 2^X,2^\tau\rangle$, which is a refinement of the topological space $\langle 2^X,2^{2^X}\rangle$ with the closure

(2) $\quad$ $M\in\overline{\alpha} \Leftrightarrow
\exists L\in\alpha: L\cap M\ne\emptyset$.

**NO, it doesn't work**: $\overline{\underset{i}\bigcap \alpha_i}\subseteq\underset{i}\bigcap \overline{\alpha_i}$ when $M\in\overline{\alpha} \Leftrightarrow
\exists L\in\alpha: \overline{L}\cap \overline{M}\ne\emptyset$. But not the opposite.

I will try to repair however.