The starting point of this question is a slight reformulation of the $T_0$ separation axiom: A topological space $(X,\tau)$ is $T_0$ if for all $x\neq y\in X$ there is a set $U\in \tau$ such that $$\{x,y\}\cap U \neq \emptyset \text{ and } \{x,y\}\not\subseteq U.$$

Given a cardinal $\kappa \geq 2$, we say that a space $(X,\tau)$ is $T^{\kappa}_0$ if for all subsets $S\subseteq X$ with $|S|=\kappa$ there is a set $U\in \tau$ such that $U$ "splits" $S$, or more formally $$S\cap U \neq \emptyset \text{ and } S\not\subseteq U.$$ Obviously, if $\lambda\geq \kappa\geq 2$ and if $(X,\tau)$ is $T^\kappa_0$, then $X$ is also $T^\lambda_0$. We say, the space $(X,\tau)$ is *minimally* $T^\kappa_0$ if it is $T^\kappa_0$, but for all cardinals $\alpha<\kappa$ with $\alpha\geq 2$, the space $(X,\tau)$ is not $T^\alpha_0$.

**Question.** Given cardinals $\lambda\geq\kappa\geq 2$, is there a topological space $(X,\tau)$ such that $|X|=\lambda$ and $(X,\tau)$ is minimally $T^\kappa_0$?