existence of minimal $T_0$topology with special properties Let $X$ be an infinite set and let $τ$ be a $T_0$-topology on $X$ that has no open finite subset and has no finite closed subset except empty set. Does $τ$ contain a $T_0$-topology on $X$ that is minimal with respect to $⊆$ ?
 A: Let $Y$ be any uncountable set, and let $X=Y\times\Bbb Z$. Let $\Phi$ be the set of finite partial functions $\varphi:Y\to\Bbb Z$, and for each $\varphi\in\Phi$ let
$$U_\varphi=\{\langle x,n\rangle\in X:x\notin\operatorname{dom}\varphi,\text{ or }x\in\operatorname{dom}\varphi\text{ and }n<\varphi(x)\}\;.$$
Let $\tau=\{\varnothing\}\cup\{U_\varphi:\varphi\in\Phi\}$; $\tau$ is a $T_0$ topology on $X$, and every non-empty open or closed set in $X$ is infinite. 

For an informal description, each non-empty open set $U$ in $X$ contains $(Y\setminus F)\times\Bbb Z$ for some finite (possibly empty) $F\subseteq Y$, and for each $x\in F$ there is an $n_x\in X$ such that $\langle x,k\rangle\in U$ iff $k<n_x$. If $U=U_\varphi$, $F=\operatorname{dom}\varphi$, and $n_x=\varphi(x)$ for each $x\in F$.

It’s useful to note that for any $\varphi,\psi\in\Phi$, $U_\varphi\subseteq U_\psi$ iff $\operatorname{dom}\varphi\supseteq\operatorname{dom}\psi$, and $\varphi(x)\le\psi(x)$ for all $x\in\operatorname{dom}\psi$.
Suppose that $\tau_0\subseteq\tau$ is minimal $T_0$. For each non-empty $U\in\tau_0$ let
$$\tau_U=\{W\in\tau_0:U\subseteq W\text{ or }W\subseteq U\}\;;$$
it’s not hard to check that $\tau_U$ is a $T_0$ topology on $X\times X$, so $\tau_U=\tau_0$, and $\langle\tau_0\setminus\{\varnothing\},\subseteq\rangle$ is an uncountable chain.
Let $\Phi_0=\{\varphi\in\Phi:U_\varphi\in\tau_0\}$, and for $\varphi,\psi\in\Phi_0$ write 
$$\begin{align*}
\varphi\preceq\psi&\text{ iff }U_\varphi\subseteq U_\psi\\
&\text{ iff }\operatorname{dom}\varphi\supseteq\operatorname{dom}\psi,\text{ and }\varphi(x)\le\psi(x)\text{ for all }x\in\operatorname{dom}\psi\;;
\end{align*}$$
clearly $\langle\Phi_0,\preceq\rangle$ is an uncountable chain. For each finite $F\subseteq X$ there are only countably many $\varphi\in\Phi$ such that $\operatorname{dom}\varphi=F$, so there is an uncountable $\Phi_1\subseteq\Phi_0$ such that $\operatorname{dom}\varphi\ne\operatorname{dom}\psi$ whenever $\varphi,\psi\in\Phi_1$ and $\varphi\ne\psi$. But then $\langle\{\operatorname{dom}\varphi:\varphi\in\Phi_1\},\supseteq\rangle$ is an uncountable chain of finite sets, which is absurd. Thus, $\tau$ does not contain a minimal $T_0$ topology.
