I have been reading the following article. I have a question in Lemma 2.3 about the closed sets $\mathcal{A}$ and $\mathcal{B}$ that are presented.In summary, my question is the following:
A generalized arc joining $p$ and $q$ in a topological space $Z$ is a subcontinuum $ \alpha $ such that $p, q \in \alpha$ and each point $z\in \alpha \setminus \{p,q\}$, separates $p$ and $q$ in $\alpha$.
Given $A,B\in 2^X$, with $A\subsetneq B$, an order arc in $2^X$, from $A$ to $B$ is a subcontinuum $\alpha$ of $2^X$ such that $A\subset C\subset B$ for each $C\in\alpha$ and, for every $C,D\in\alpha$, either $C\subset D$ or $D\subset C$.
The lemma reads as follows:
Let $X$ be a Hausdorff continuum, $A,B\in 2^X$ and let $\alpha$ be an order arc in $2^X$, from $A$ to $B$. Then $\alpha$ is a generalized arc joining $A$ and $B$ in $2^X$.
Proof. It is enough to show that for each $C\in \alpha\setminus \{A,B\}$, $C$ separates $A$ and $B$ in $\alpha$. Let $\mathcal{A}=\{D\in \alpha:D\subset C\}$ and $\mathcal{B}=\{D\in \alpha:C\subset D\}$. It is east to show that $\mathcal{A,B}$ are closed subsets of $\alpha$. By the definition of order arc, we have $\alpha =\mathcal{A\cup B}$, $A\in \mathcal{A}\setminus \{C\}$, $B\in \mathcal{B}\setminus \{C\}$ and $\mathcal{A\cap B}=\{C\}$. This proves that $C$ separates $A$ and $B$ in $\alpha$.
My question is why sets $\mathcal{A}$ and $\mathcal{B}$ are closed. We are considering the Vietoris topology over $2^X$. In the first case, for $\mathcal{A}$, we see that $ C $ is closed at $\alpha$, then by the Vietoris topology we know that $\langle A\rangle=\{ F\in 2^X: F\subset A \}$ is closed when $A$ is closed but in the case of $\mathcal{B}$ I don't know how to argue why it is closed.
I would appreciate any help in advance.
$\mathcal{B}= \{D \in \alpha\mid C \subseteq D\} = \alpha \cap \left(\bigcap_{x \in C} [\{x\}] \right)$, which is closed in $\alpha$, where $[F] = \{A \in 2^X\mid A \cap F \neq \emptyset\}$ is the other type of Vieteris subbasic set (it's open if $F$ is open, and closed if $F$ is closed, as is well-known and part of the definition and folklore of this topology).
So we've written $\mathcal{B}$ as an intersection of closed sets, relativised to $\alpha$, so it's indeed closed in $\alpha$. Interestingly, $C$ can be any non-empty set and the condition $C \subseteq D$ is always a "closed condition"...