Is there a connected $T_1$ anticompact topological space? Now cross-posted to Mathoverflow.
A topological space is a called anticompact if all its compact subsets are finite.

Question 1: Is there a (non-trivial) connected $T_1$ anticompact space?

Obviously the singleton and the empty space satisfy these properties.  So let's assume such a space $X$ has at least two points.  Here are some observations:
(1) $X$ must be infinite.  Because finite $T_1$ spaces are discrete, such a space would not be connected.
(2) $X$ cannot be compact.
(3) $X$ must be totally path disconnected, because anticompact $T_1$ spaces are totally path disconnected.  (The image of a continuous map $[0,1]\to X$ is compact, connected and $T_1$, hence finite and $T_1$, so it would be discrete, and connected. Thus it must a single point.)
(4) $X$ cannot be sequential, as a consequence of this. In particular, $X$ cannot be first countable.

Answer (from Ulli): Yes.  Any uncountable set with the cocountable topology is connected and anticompact.  It is also $T_1$, but not $T_2$.
So updating the question:

Question 2: Is there a (non-trivial) connected Hausdorff anticompact space?


I think the following would be an example.  Take $X=[0,1]$. Let $\tau_1$ be the usual Euclidean topology on $X$ and take on $X$ the topology $\tau$ generated by $\tau_1$ and the cocountable topology.  By using the Lindelof property for $\mathbb{R}$ it's not too difficult to show that the topology is described by
$$\tau=\{U\setminus A: U\in\tau_1 \text{ and } A\subseteq X \text { is countable}\}.$$
Every countable subset of $X$ is closed, and is discrete since all its subsets are also countable, hence closed.  So if $A$ is an infinite subset of $X$, it cannot be compact; otherwise a countably infinite subset of it would also be compact (as closed in a compact set) and discrete, which is impossible.  This shows that $X$ is anticompact.
$X$ is Hausdorff.
To show that $X$ is connected, suppose $X=Y\cup Z$ with $Y$ and $Z$ nonempty disjoint open sets.  Take a point $x\in Y$.  There is a Euclidean nbhd $U$ of $x$ such that $Y$ contains all the points of $U$ except countably many of them (where countable = finite or countably infinite).  Any point $z\in U\setminus Y$ would have to be contained in $Z$.  And since $Z$ is open in $X$, $U\cap Z$ would be a countable nbhd of $z$, which is impossible.  This shows that both $Y$ and $Z$ are open in the Euclidean topology.  Again that is impossible since $[0,1]$ is connected in its usual topology.
I would appreciate if you could confirm that the example above is correct.

This leads me to the revised question:

Question 3: Is there a countably infinite connected $T_1$ (or $T_2$) anticompact space?

I can't find an example.  The Arens-Fort space or the Appert space are anticompact (and not first countable) but not connected.  Countably infinite connected examples that I know of, like the irrational slope topology or the Golomb space, are first countable, hence not anticompact.
 A: Here is a partial answer to question 3, namely an example of a connected, anticompact T1-topology on the (positive) integers:
Let  $\mathfrak{F}$ be a free ultrafilter on $\mathbb{N}$ and $\tau := \mathfrak{F} \cup \{\emptyset\}$ the ultrafilter topology on $\mathbb{N}$.
Since $\mathfrak{F}$ is free, for each $n \in \mathbb{N}$, $\mathbb{N} \setminus \{n\} \in \mathfrak{F}$, hence $\{n\}$ is closed. Thus, the topology is T1.
The intersection of any two non-empty, open sets is non-empty again. Hence the topology is connected.
It is anticompact:   
Let $A$ be an infinite subset of $\mathbb{N}$. Then there exists $U \in \mathfrak{F}$ such that $A \setminus U$ is infinite.  
[$A$ is the disjoint union of two infinite subsets $S, T$. If $S \in \mathfrak{F}$ or $T \in \mathfrak{F}$, we are done. Otherwise,
$\mathbb{N} \setminus S  \in \mathfrak{F}$ and
$\mathbb{N} \setminus T  \in \mathfrak{F}$.
Hence
$\mathbb{N} \setminus A =  \mathbb{N} \setminus (S \cup T) = (\mathbb{N} \setminus S) \cap (\mathbb{N} \setminus S) \in \mathfrak{F}$.] 
Now, $A \setminus U$ is discrete. 
[For any $x \in A \setminus U$,
$\{x\} =  (U \cup \{x\}) \cap (A \setminus U)$
is open in $A \setminus U$.] 
Furthermore, $A \setminus U \notin \mathfrak{F}$.
Hence its complement is open, hence $A \setminus U$ is closed in $\mathbb{N}$ and thus in $A$. Therefore $A$ cannot be compact.
Note:
Of course, this topology is far from being Hausdorff.
A: As answered in the cross-post to mathoverflow, a paper by T. Banakh and Y. Stelmakh provides examples of countably infinite connected Hausdorff anticompact spaces.  Their examples are even strongly rigid.
