A space $X$ is called anti-compact if any compact subset of $X$ is finite. It is known that uncountable space with co-countable topology is an example of a $T_1$ anti-compact space that is not Hausdorff.
I am looking for other examples. Is there any one can give me another example of an anti-compact space that is $T_1$ but not Hausdorff?
At the moment I see a couple of easy ways to build more examples.
Let $Y$ be an uncountable space with the co-countable topology, let $Z$ be a discrete space, and let $X=Z\times Y$. Clearly $X$ is $T_1$ but not Hausdorff. Suppose that $A\subseteq X$ is infinite, and for each $z\in Z$ let $A_z=A\cap\big(\{z\}\times Y\big)$. If $\{z\in Z:A_z\ne\varnothing\}$ is infinite, then $A$ is clearly not compact, so $A_z$ is infinite for some $z\in Z$. But then $A_z$ is a closed, non-compact subset of $A$, so $A$ is not compact. It follows that $X$ is anticompact. Moreover, the maximum cardinality of a family of pairwise disjoint, non-empty open sets in $X$ is $|Z|$, so for a given $Y$ we really do get different spaces when we change $|Z|$.
Alternatively, we could start with any non-discrete anticompact Hausdorff space $X$; there are three in this answer by bof, and I’ll add one more that is similar to the last two in bof’s answer:
Let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$, and let $X=\{\mathscr{U}\}\cup\Bbb N$; then $$\tau=\wp(\Bbb N)\cup\big\{\{\mathscr{U}\}\cup U:U\in\mathscr{U}\big\}$$ is an anticompact Hausdorff space that is not discrete.
Let $p$ be any limit point of $X$, and let $q$ be a point not in $X$. Let $\mathscr{N}$ be the family of open nbhds of $p$. Let $Y=X\cup\{q\}$; the topology on $Y$ is
$$\{Y\}\cup\tau\cup\big\{\{q\}\cup(N\setminus\{p\}):N\in\mathscr{N}\big\}\;.$$
In other words, the open nbhds of $q$ are identical to those of $p$, except that the point $p$ is replaced by the point $q$. Clearly $Y$ is no longer Hausdorff, but it is still $T_1$ and anticompact.
The first of bof’s examples has no isolated points, so we can do even better: it’s the real line with the topology generated by the usual open sets and the co-countable sets. Let $\langle X,\tau\rangle$ denote this space. Let $Y=X\times\{0,1\}$, and let
$$\mathscr{B}=\big\{(U\times\{0,1\})\setminus F:U\in\tau\text{ and }F\subseteq Y\text{ is finite}\big\}\;;$$
then $\mathscr{B}$ is a base for an anticompact $T_1$ topology on $Y$ such that for each $y\in Y$ there is a $y'\in Y$ such that $y$ and $y'$ do not have disjoint open nbhds: if $y=\langle x,i\rangle$, then $y'=\langle x,1-i\rangle$.
