According to the Scholium 1.1.4 in Part C of Peter's Johnstone ,,Sketeches of the Elephant'' the free frame $F(A)$ over a set $A$ can be realized as $\mathcal{O}(X)$ for suitable topological space $X$. The space $X$ can be constructed as $KA^{op}$ (the family of finite subsets of $A$ with the inverse order) with the Alexandroff topology (where a set is open iff it is lower set) but also $X$ can be realized as $S^A$ where $S$ is the Sierpiński space. It is known that two topological spaces $X,Y$ are homeomorphic iff the frames $\mathcal{O}(X)$ and $\mathcal{O}(Y)$ are isomorphic provided these two spaces are sober. Since the cardinality of $KA$ is less than that of $S^A$ these two spaces clearly are not homeomorphic. Therefore one of them can not be sober. Somewhere I have read that the product of sober spaces is again sober.
Could someone point me some reference for this result?
If it is so, then $KA^{op}$ cannot be sober:
However I would also like to see a direct proof of the fact that $KA^{op}$ is not sober.
To prove that $KA^{op}$ is not sober (for $A$ infinite), just observe that $KA^{op}$ is irreducible: any two basic open sets have nonempty intersection, since you can take the union of the corresponding finite subsets of $A$. But $KA^{op}$ does not have a generic point, since there is no largest finite subset of $A$.