Before proving the fact you want, we'll need the notion of quasicomponents and some basic propositions about it. In these terms, you are just asking if distinct points are in distinct quasicomponents. Let $X$ be a topological space. Given $x,y \in X$, define $x \sim y$ if $X$ cannot be written as a disjoint union of open sets $U$ and $V$ containing $x$ and $y$, respectively. It is straightforward to verify that $\sim$ is an equivalence relation. The equivalence classes are called the quasicomponents of $X$. It's easy to prove that the quasicomponent of a point $x$ is the intersection of all closed-open subsets of $X$ containing $x$.
In every topological space, the component $C$ of a point $x$ is contained in the quasicomponent $Q$ of the point $x$. In fact, if $F$ is a closed-open set containing $x$, then $X = F \cup (X-F)$ is a separation of $X$. Since $C \cap F \ne \emptyset$, it follows that $C \subseteq F$. Thus we have $C \subseteq Q$.
I'll prove now, based on Engelking's proof in the book General Topology, that in every compact Hausdorff space, components and quasicomponents coincide. Let $C$ and $Q$ as above. We just need to prove that the quasicomponent $Q$ is connected. Then will follow that $ Q = C$. Suppose that $Q = X_1 \cup X_2$, where $X_1, X_2$ are two disjoint closed subsets of the space $Q$. Then $X_1$ and $X_2$ are closed in $X$, since $Q$ is closed in $X$. By normality of compact Hausdorff spaces, there exist disjoint open subsets $U,V$ of $X$ containing $X_1, X_2$, respectively. Hence, we have $Q \subseteq U \cup V$ and, by compactness, there exist closed-open sets $F_1, \ldots, F_k$ such that
$$Q \subseteq \bigcap_{i=1}^k F_i \subseteq U \cup V.$$
$F = \bigcap_{i=1}^k F_i$ is clearly closed-open. Since $ \overline{U \cap F} \subseteq \overline{U} \cap F = \overline{U} \cap (U \cup V) \cap F = U \cap F$, the intersection $U \cap F$ is also closed-open. As $x \in U \cap F$, we have $Q \subseteq U \cap F$ and $X_2 \subseteq Q \subseteq U \cap F \subseteq U$. It follows that $X_2 \subseteq U \cap V = \emptyset$, which shows that the set $Q$ is connected.
Now it's simple to prove your statement. Let $B$ be a Stone Space. Since $B$ is compact Hausdorff, quasicomponents coincide with components. $B$ is totally disconnected, so the quasicomponent of a point $a \in B$ is $\{a \}$. If $b \in B$ is a different point, then $b$ isn't in the quasicomponent of $a$. Thus there exist disjoint open sets $U,V$ containing $a,b$, respectively, such that $B = U \cup V$.
Added This was motivated by Pete's answer. Actually, to prove that a locally compact Hausdorff totally disconnected space $X$ is zero-dimensional, we just need what I've proved for compact Hausdorff spaces. Indeed, let $x \in X$ and $U$ an open set containing $x$. Since $X$ is regular, there is an open set $V$ containing $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$. Also, $\overline{V}$ is a totally disconnected space. Using the fact that quasicomponents and components coincide in compact Hausdorff spaces, we have that the quasicomponent of $x$ in $\overline{V}$ is $\{x\}$. Now, the compactness of $\overline{V}$ guarantees that there are closed-open sets $F_1, \ldots, F_k$ such that $x \in F_1 \cap \cdots \cap F_k \subseteq V$. Let $F$ be $\bigcap_{i=1}^k F_i$. Then $F \,$ is an open set in $V$ and since $V \, $ is open in $X$, $F \, $ is open in $X$. Also, $F \,$ is compact, since each $F_i$ is a closed subset of the compact space $\overline{V}$, and then $F\,$ is closed in $X$. We conclude that for each open set $U$ containing $x$, there is a closed-open set $F \ $ such that $x \in F \subseteq U$. Thus $X$ is zero-dimensional.
If $A$ is countable, then $\{u\} = \bigcap_{a \in u} \{u \in Ult(A) : a \in u\}$. So we should have $\{\{u\} : u \in Ult(A)\} \subseteq B$.
If $A$ is allowed to be uncountable, I think the following would be a counterexample. Let $A$ be freely generated by uncountably many generators $\{a_i\}_{i \in I}$. As you mentioned, the clopens of the Stone space $C$ are given by the elements of $A$ in the sense that for $a \in A$ we have a clopen $[a] = \{u \in Ult(A): a \in u\}$. Any non-empty element $V$ in the $\sigma$-algebra $B$ generated by the Stone space $C$, is a countable Boolean combination of clopens. Every element in $A$ is given by some finite Boolean combination of generators. In particular there will be only countably many generators that are relevant for $V$. Since we had uncountably many generators, we can find some generator $a$ that does not occur anywhere in $V$. So there will be at least two distinct ultrafilters in $V$, one that contains $a$ and one that contains $\neg a$. We can do this for every (non-empty) element of the $\sigma$-algebra, so there will be in fact no singletons in the $\sigma$-algebra generated by the Stone space of $A$.
Best Answer
$S(B)$ contains all countably many fixed ultrafilters $\mathcal{U}_n=\{A \subseteq \omega: n \in A\}$ (for any fixed $n \in \omega$)(also called "principal ultrafilters") but many many more ($2^{2^{\aleph_0}}$ many!) non-fixed, free ultrafilters, that have empty intersection, and no common point. These are hard to imagine (you cannot give a constructive example of one) but Zorn's lemma implies that they exist. The fixed ultrafilters as a subspace of $S(B)$ form a countable discrete set. The Stone space is homeomorphic to what topologists call the Cech-Stone (or Stone-Cech) compactification (see Wikipedia e.g.) of the integers and the fixed ultrafilters "are" in that compactfication the original integers. The other ultrafilters are the compactifying points we add.
So your picture is incomplete. K.P. has a nice intro to Stone duality here, as an extra resource.