I am trying to prove that an algebra of the ultrafilter monad $\mathrm{lim} : \beta X \to X$ is equivalent to a (compact hausdorff) topological space. I'm stuck at proving that it is a topological space in the first place - that a union of opens (defined by $U$ is open $:= \forall F \in \beta X, \, U \in F \implies \mathrm{lim} F \in U$) is open.
Suppose that $U_i$ is open for every $i$ in some arbitrary index set $I$. Then from $\cup_{i\in I} U_i \in F$, if I were to prove some $U_i \in F$, then by openness of $U_i$ we get $\mathrm{lim} F \in U_i \subset \cup_{i\in I} U_i$, proving the union is open.
The problem is that I can't find a way to get some $U_i \in F$. The contravariance of implication makes this definition of opens seem strange. I used the upwards closed hypothesis to prove that the intersection of two opens is open, when it seems like I should be using the finite intersection property to prove that, and upwards closure for this. Since ultrafilters are quite unconstructive, I tried assuming that each $U_i \not \in F$, but it seems like I need arbitrary intersection to prove that $\varnothing \in F$, deriving a contradiction.
Direction towards proving this, or corrections to an error I've made is preferred over a complete proof.