Is the following theory/proof correct?
Let $X$ be a nonempty set.
A family of subsets $\mathcal K$ of $X$ is said to be a family of filaments for $X$ if the following conditions hold:
$\tag 1 \displaystyle \bigcup_{K \in \mathcal K} K = X$
$\tag 2 \text{If } K,L \in \mathcal K \text{ Then } K \cap L = \emptyset \text{ or } K \subset L \text{ or } L \subset K$
If $x \in X$ let $\mathcal K_x = \{K \in \mathcal K \mid x \in K\}$ and define
$\quad \displaystyle \mathcal G(x) = \bigcup_{K \in \mathcal K_x}K$
Proposition: The assignment $x \mapsto \mathcal G(x)$ is a (partition/equivalence class) projection mapping.
Proof
Firstly, note that $x \in \mathcal G(x)$.
It remains to show that if $\mathcal G(x) \cap \mathcal G(y) \ne \emptyset$ then $\mathcal G(x) = \mathcal G(y)$.
If $z$ belongs to the intersection then we have filaments $K$ and $L$ such that $x,z \in K$ and $y,z \in L$. We conclude by $\text{(2)}$ that both $x$ and $y$ belong to a common filament $J$ (equal to either $K$ or $L$). But once we have $J$, by using $\text{(2)}$ again we see that $\mathcal G(x)$ is the union of $J$ with all superset filaments, and the same is true for $\mathcal G(y)$. So they are indeed equal. $\quad \blacksquare$