11.22 Lemma, from B. A. Davey, H. A. Priestley, Introduction to lattices and order,
let $L$ be a bounded distributive lattice with dual space $(X:=\mathcal{I}_p(L), \subseteq, \tau)$ and $X_a = \{I \in X \: | \: a \notin I\}$, then:
(i) the clopen subsets of $X$ are finite unions of sets of the form $X_b \cap (X \setminus X_c)$ for $b, c \in L$
(ii) the clopen down-sets of $X$ are exactly the sets $X_a$ for $a \in L$.
I've proved (i) with ease, since if $U$ is a clopen set of $X$, it is open, so it is union of members of the base $\mathcal{B} = \{X_b \cap (X \setminus X_c) \: | \: b, c \in L\}$ and since $U$ is a closed set in a compact Hausdorff space it is compact, so that union is finite.
Any tips about how to prove (ii)? $\:$If $I$ is a clopen down-set of $X$, then $I$ it is a finite union of elements (i), like $I = (X_{b_1}\cap(X \setminus X_{c_1}))\cup \dots \cup ((X_{b_k} \cap (X \setminus X_{c_k}))$ and I imagine that I need to write it as $X_a$ for an $a$ s.t. it is join and meet of the elements $b_1, \dots , b_k, c_1, \dots , c_k$?
Consider a prime ideal $I$ in a clopen downset $D$. The downset generated by $I$ is the intersection of all the sets $X_{a}$ such that $a \notin I$. Because $D$ is compact, there is a finite subintersection which is a subset of $D$. A finite intersection of sets of the form $X_{a}$ is a set of the form $X_{a}$, so for each $I$ in $D$ there is some $a$ such that $I \in X_{a} \subseteq D$. The union of all sets $X_{a} \subseteq D$ is therefore equal to $D$. Compactness yields a finite subunion, and a finite union of sets of the form $X_{a}$ is again a set of the form $X_{a}$.