Situation: The Vietoris functor $\mathbb{V}$ is an endofunctor on the category of topological spaces. Let $\mathcal{X}$ be a topological space, then $\mathbb{V}\mathcal{X}$ is the set of closed subsets of $\mathcal{X}$ topologized by the subbase $$ \Box U := \{ V \in \mathbb{V}\mathcal{X} \mid V \subseteq U \}, \qquad \Diamond U := \{ V \in \mathbb{V}\mathcal{X} \mid U \cap V \neq \emptyset \}, $$ where $U$ ranges over the opens of $\mathcal{X}$.
Question: Suppose $\mathcal{X}$ is compact and sober. Is $\mathbb{V}\mathcal{X}$ compact and sober?
Thoughts: It is known that $\mathbb{V}$ preserves compactness. I have never found a proof that states $\mathbb{V}$ preserves sobriety, so I don't think this is true. So I wonder whether the additional compactness "helps" the preservation of sobriety.