Let $X$ be a set, and the power set of $X$ be $\mathcal P(X).$
For $\mathcal B \subset \color{red}{\mathcal P(X)},$ does \begin{align} &\quad \ \left\{ U\subset X \mid \exists \Lambda : \mathrm{index \ set}, \exists \{U_\lambda\}_{\lambda \in \Lambda}\subset \mathcal B\ \ ;\ U=\cup_{\lambda\in \Lambda} U_\lambda\right\}\\ &=\{ U\subset X \mid \forall x\in U, \exists B\in \mathcal B \ ; \ x\in B \subset U \} \end{align} hold ?
I expect it holds. I proved LHS$\subset $RHS, but I'm not sure my proof of LHS$\supset $RHS is correct.
My proof.
Let $U\in $ RHS.
Then, $\forall x\in U, \exists B_x \in \mathcal B \ ; \ x\in B_x \subset U $ holds.
Each $B_x$ belongs to $\mathcal B$ i.e., $\{ B_x \}_{x\in \color{red}U}\subset \mathcal B$, and $U=\cup_{x\in \color{red} U} B_x$ holds, so $U\in $LHS.
I'm not sure this is correct, since I don't know whether we can see $\color{red}U$ as index set.
$\color{red} U$ is defined as a set, but can we see $\color{red} U$ as an index set ?
Any set can be used as an index set. Index sets are not special types of sets, they're just ordinary sets that are being used in a special way: to index a family of other sets.
That said, your proof is not quite right, because you have defined a set $B_x$ for every $x\in U$, not for every $x\in X$. So your index set should be $U$, not $X$.