Following Bankston – The total negation of a topological property, a topological space is called anticompact if all its compact subsets are finite. The linked MSE post above has two examples:
Example 1: The topology on $\mathbb{R}$ generated by the usual Euclidean topology together with the cocountable topology is an uncountable space that is connected anticompact and Hausdorff.
Example 2: Given a free ultrafilter $\mathfrak{F}$ on $\mathbb{N}$, take the ultrafilter topology $\tau = \mathfrak{F} \cup \{\emptyset\}$ on $\mathbb{N}$. This is a countably infinite, connected anticompact space. It is $T_1$, but not $T_2$.
Does there exist a countably infinite connected anticompact space that is Hausdorff?
Note that such an example would have to be totally path disconnected. Also it cannot be sequential, as a consequence of this answer to Is a space where only finite subsets are compact sets always discrete?. In particular, it cannot be first countable.
Best Answer
Such an example has been constructed by Banakh and Stelmakh. More precisely, they constructed an anticompact countable connected Hausdorff space which is Brown and strongly rigid.