Let $\mathcal A$ be any family of subsets of $X$. Let:
$C(\mathcal A) := \{\ \bigcup \mathcal B: \mathcal B \subseteq \mathcal A \text { is up-directed by inclusion} \}$
$I(\mathcal A) := \{ \bigcap \mathcal B: \mathcal B \subseteq \mathcal A\}$
Is it guaranteed that $C(I(\mathcal A)) = I(C(\mathcal A))$?
If not, can it at least be guaranteed provided that $C(\mathcal A)=\mathcal A \cup \{A^\complement : A \in \mathcal A\} \cup \{\emptyset, X\}$ ?
Ok, $C(I(\mathcal A)) = I(C(\mathcal A))$ for any $X, \mathcal A \subseteq 2^X$:
$C(I(\mathcal A))$ is the coarsest convexity containing $\mathcal A$. Elementarily, $I(C(\mathcal A))$ is a closure and$I(C(\mathcal A)) \subseteq C(I(\mathcal A))$, thus if $I(C(\mathcal A))$ can be proven to be a convexity - it must equal $C(I(\mathcal A))$.
$I(C(\mathcal A))$ is a convexity iff it's closure operator is finitary - that is, if for any $B \subseteq X, p \in cl_{I(C(\mathcal A))}(B)$ there is finite $C \subseteq B$ such that $p \in cl_{I(C(\mathcal A))}(C)$. $[1]$
$$cl_{I(C(\mathcal A))}(B)= \bigcap \{A \in I(C(\mathcal A)): B \subseteq A\} = \bigcap \{\bigcup \mathcal B : \text{up-directed } \mathcal B \subseteq \mathcal A, B \subseteq \bigcup \mathcal B\}$$
If $D \subseteq B$ is finite, then for any up-directed $\mathcal B \subseteq \mathcal A: D \subseteq \bigcup \mathcal B$ there must exist $A \in \mathcal B: D \subseteq A$. Thus:
$$cl_{I(C(\mathcal A))}(D) = \bigcap \{A \in \mathcal A: D \subseteq A\}$$
If for any finite $D \subseteq B$ there is $A \in \mathcal A: D \subseteq B,\ p \notin B$ then it is easy to show that one can construct an up-directed family $\mathcal {B \subseteq A}$ such that $B \subseteq \bigcup \mathcal B$, $p \notin \bigcup \mathcal B$. Thus $cl_{I(C(\mathcal A))}(B)=\cdots$ $\cdots =\bigcup \{cl_{I(C(\mathcal A))}(D): \text{finite} \ D \subseteq B\}$, hence $cl_{I(C(\mathcal A))}$ is finitary, hence $C(I(\mathcal A))=I(C(\mathcal A))$.
$[1]$ Van de Vel, Theory of convex structures, theorem $1.3$ on page $5$