Boolean Algebra with decomposition property

Question

1. Does there exist a (countable) Boolean algebra $(B,\bigcup, \bigcap, 0, 1)$ with the following property:

$\forall A\in B\setminus \{0\}$ there exists $A_1,A_2\in B$ such that $A_i\neq A$, $A_1 \bigcup A_2 = A$ and $A_1 \cap A_2=0$.

(e.g. in the uncountable case: take the Boolean algebra consisting of all infinite subets of the natural numbers)

2. Does such a property have a (well established) name?

Answer 1

What you're looking for is a Boolean algebra $\mathbf{B} = \langle B, \wedge, \vee,',0,1 \rangle$ without join-irreducible elements.

Definition. In any lattice (in particular, a Boolean algebra) an element $x$ is said to be join-irreducible if $x = a$ or $x = b$, whenever $x = a \vee b$.

In a lattice with $0$, an atom is an element $x$ such that $y = 0$, whenever $y<x$.

Lemma. If $\mathbf{B} = \langle B, \wedge, \vee,',0,1 \rangle$ is a Booelan algebra and $j \in B$ is a join-irreducible element, then $j$ is an atom.
Proof. Let $0 \leq x < j$. We ought to prove that $x=0$. We have $$j = x \vee j = (x \vee j) \wedge (x \vee x') = x \vee (j \wedge x').$$ Since $j$ is join-irreducible and $x < j$, it follows that $j = j \wedge x'$, that is, $j \leq x'$, whence $$x = x \wedge j \leq x \wedge x' = 0.$$

So, rephrasing the first paragraph, you're looking for an atomless Boolean algebra.
The answer is yes, there are atomless Boolean algebras and in particular, countable ones. See, for example, this Wikipedia article.