Is this a basis for the Scott topology on the collection of open sets of a topology?

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If $\mathcal{O}(X)$ is the collection of open sets in the topological space $X$ ordered by inclusion and $K\subseteq X$ is a compact subset of $X$ then it is easy to see that:

$$\mathscr{U}_K = \{O\in\mathcal{O}(X) : K\subseteq O\}$$

is open in the Scott topology on $\mathcal{O}(X)$. If $X$ is finite then the collection $\{\mathscr{U}_K:K\ \mbox{compact in}\ X\}$, is clearly a basis for this topology.

My question is: Is this collection always a basis for the Scott topology on $\mathcal{O}(X)$?