Let $E$ be the set whose subsets only we’ll be talking of. Now complementation is a bijection on $\mathcal{P} (E)$. Hence the following two sets seem intuitively equal: \begin{align}& \{ x\in E: \forall X(X\in \mathcal{C} \rightarrow x\in X)\} \text{; and,}\\ & \{ x\in E: \forall X(X’\in \mathcal{C} \rightarrow x\in X’)\}.\\ \end{align}
Here, $\mathcal{C} \subset \mathcal{P} (E)$.
How to show formally that they are indeed equal?
($X’$ means relative complement of $X$ in $E$.)
To show that the two sets are equal is the same as showing for arbitrary $x, \mathcal C, E$:
$$ \big[ \forall X(X\in \mathcal{C} \rightarrow x\in X) \bigr] \iff \big [ \forall X(X’\in \mathcal{C} \rightarrow x\in X’) \bigr] $$
Let's do the $\Longleftarrow$ direction of this; the other is similar. So we're assuming $$ \tag{a} \forall X(X'\in \mathcal{C} \rightarrow x\in X')$$ Now we're given a particular $X$ and we must then show $$ \tag{g} X\in\mathcal{C} \to x\in X. $$
The $X$ in (a) is a dummy variable, so its name does not matter, and the assumption is the same as $$ \tag{b} \forall Y(Y'\in \mathcal{C} \rightarrow x\in Y')$$ Now instantiate (b) by substituting $X'$ for $Y$ in the scope of $\forall Y$, which gives us $$ X''\in\mathcal{C} \to x \in X'' $$ But -- at least under the tacit assumption that $X\subseteq E$ -- we have $X''=X$ so this is the same as (g), which was what we needed to prove. We're done!