Assume that $\mathcal{A} \neq \emptyset$ is an arbitrary family of subsets of $X$, such that $(\mathcal{A}, \subseteq)$ is a total ordering. I want to prove that the intersection of all members of $\mathcal{A}$ is also a member of $\mathcal{A}$; i.e that $\bigcap\mathcal{A} \in \mathcal{A}$.
For all finite $\mathcal{A}$ this can be done by induction. Note that for all $A,B \in \bigcap{A}$, $A \subseteq B$ and $A \cap B = A$, or $B \subseteq A$ and $A \cap B = B$; in any case, $(A \cap B) \in \mathcal{A}$. Base case: $\bigcap\{A\} = A$ and $A \in \{A\}$. Inductive step: assume true for all families of cardinality $n \in \mathbb{N} \ge 1$. Given $|\mathcal{A}| = n + 1$ let $\mathcal{B} = \mathcal{A} \setminus \{A\}$ for some $A \in \mathcal{A}$; then $|\mathcal{B}| = n$, so $\bigcap\mathcal{B} \in \mathcal{B}$ and thus $\bigcap\mathcal{B} \in \mathcal{A}$. Consequently $\bigcap\mathcal{A} = A \cap (\bigcap\mathcal{B}) \in \mathcal{A}$.
Now I'm trying to prove it for all infinite $\mathcal{A}$ too. I've tried to prove it directly, contrapositively, and by contradiction (assuming $\bigcap\mathcal{A} \notin \mathcal{A}$), but I am stuck. Any hint would be greatly appreciated.
It has been pointed out in the comments that the statement is not generally true for infinite families, and so cannot be proven.