Let $(X, \mathcal{T})$ be a topological space, and $\mathcal{T} \cap \mathcal{T}^C$ be a topology, where $\mathcal{T}^C$ denotes closed subsets. That is, the clopen subsets form a topology.
Are all connected components of $(X, \mathcal{T})$ open?
Background:
Theorem
If the connected components of $(X, \mathcal{T})$ are all open, then $\mathcal{T} \cap \mathcal{T}^C$ is a topology.
Proof
Clearly the clopen subsets are closed under finite intersections. Let $U \in \mathcal{T} \cap \mathcal{T}^C$. Then $U \cap C \in (\mathcal{T}|C) \cap (\mathcal{T}^C|C)$ for any connected component $C$. Since $C$ is connected, $U \cap C \in \{\emptyset, C\}$. Therefore $U$ is a union of connected components. Let $\mathcal{U} \subset \mathcal{T} \cap \mathcal{T}^C$. By the previous, $\bigcup \mathcal{U}$ is a union of connected components. It can be shown that in a space with open connected components, unions of connected components are clopen. Therefore $\bigcup \mathcal{U} \in \mathcal{T} \cap \mathcal{T}^C$. That is, $\mathcal{T} \cap \mathcal{T}^C$ is a topology. $\square$
So my question asks whether the the conditions are equivalent.
Theorem
$\mathcal{T} \cap \mathcal{T}^C$ is a topology if and only if $\partial_X \bigcup \mathcal{U} = \emptyset$ for each $\mathcal{U} \subset \mathcal{T}$ such that $\partial_X U = \emptyset$ for each $U \in \mathcal{U}$.
Best Answer
Yes: if the clopen sets form a topology, then components are open:
By the prerequisite, arbitrary unions of clopen sets are clopen. Hence, arbitrary intersections are clopen. In particular, for any $x \in X$ its quasi-component $= \bigcap \{U: x \in U \text{ is clopen in } X \}$ is clopen. Hence it is connected and therefore equals the component of $x$.