How do finite door spaces work?

Question

Recall that a door space is a topological space where every set is either open or closed (or both). A topological space is finite if it has finitely many points. I'm interested in learning about finite door spaces.

Question 0: What is an example of a finite topological space which is $T_0$ but not a door space?

Question 1: Let $n$ be a natural number. How many door topologies are there on a set with $n$ elements?

I'm interested in both the "labeled" and "unlabeled" versions of question 1.

Recall that via the specialization order, finite topological spaces are equivalent to finite preorders, i.e. quasi-(partially)ordered sets.

Question 2: Which finite posets are the specialization order of a door space?

Answer 1

A door space $X$ is $T_0$, for if $x,y\in X$ are not separated by the $T_0$ axiom then the set $\{x\}$ is neither open nor closed. A finite $T_0$ space is equivalent to a finite poset $P$ (Enumerative Combinatorics, vol. 1, second ed., Exercise 3.3). An open set corresponds to an order ideal of $P$. Thus we want to count posets on an $n$-element set such that every subset of $P$ is either an order ideal or the complement of an order ideal. Such posets can have at most one connected component $Q$ which is not a single point. Then $Q$ consists of an element $x$ at the bottom and some positive number of elements $y$ covering $x$ (i.e., $x<y$ with nothing in between), or the dual of this connected component. There is one choice if $P$ is a disjoint union of points. There are $n(n-1)$ choices if $x$ is covered by one element, but then the dual has the same form and should not be counted again. If $x$ is covered by more than one element, then there are $n(2^{n-1}-n)$ choices. This must be multiplied by 2 since the dual is different. Hence the total number of door spaces on an $n$-element set is $$ 1+n(n-1)+2n(2^{n-1}-n) = 1-n-n^2+n2^n. $$ By similar reasoning, the number of unlabelled $n$-element door topologies is $1+1+2(n-2)=2n-2$.