If clopen subsets form a topology, are connected components open? 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 $U \in \mathcal{U}$.
 A: 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 $x$.
A: This doesn't answer the question, but perhaps brings some clarity. I'll abbreviate a connected component as a component.
Definition
A space is sum-connected, if all components are open.
Theorem
A space $(X, \mathcal{T})$ is sum-connected if and only if components are locally finite.
Proof
$\implies$
Suppose $(X, \mathcal{T})$ is sum-connected. Let $x \in X$. Since components partition the space, there exists a unique component $C$ such that $x \in C$. By assumption, $C \in \mathcal{T}$. So $C$ is an open neighborhood of $x$ which intersects only one component. Therefore components are locally finite.
$\impliedby$
Suppose components are locally finite. It can be shown that components are closed, and that a union of locally finite set of closed subsets is closed. Let $C$ be a component. Since components partition the space, $X \setminus C$ is a union of components. By the previous notes, $X \setminus C \in \mathcal{T}^C$. Therfore $C \in \mathcal{T}$; i.e. $(X, \mathcal{T})$ is sum-connected.
Theorem
A space $(X, \mathcal{T})$ is sum-connected if and only if non-open components are locally finite and $\mathcal{T} \cap \mathcal{T}^C$ is a topology.
Proof
$\implies$
Suppose $(X, \mathcal{T})$ is sum-connected. By the previous proof, components are locally finite. By the proof in the question, $\mathcal{T} \cap \mathcal{T}^C$ is a topology.
$\impliedby$
Suppose non-open components are locally finite and $\mathcal{T} \cap \mathcal{T}^C$ is a topology. Let $D$ be the union of all open components. Since each component is also closed, the sets in this union are clopen subsets. By assumption, $D$ is clopen. Therefore $X \setminus D$ is clopen. Since components partition space, $X \setminus D$ is a union of components too. By assumption, these components are locally finite. Hence, we may pick one of these, and show that it is open, because the union of the others is closed. Therefore all components are open; i.e. $(X, \mathcal{T})$ is sum-connected.
What remains
The theorem above suggests that sumconnectedness is not equivalent to $\mathcal{T} \cap \mathcal{T}^C$ being a topology. However, it can still be that $\mathcal{T} \cap \mathcal{T}^C$ being a topology implies locally finite non-open components. So without a counter-example we have not yet actually answered the question.
