Finite covers of Boolean algebras by their subalgebras It is a student exercise that no group can be represented as a set-theoretic union of its two proper subgroups. The same also can be shown  for Boolean algebras. On the other hand, it's not hard to show that any infinite Boolean algebra $\mathcal{A}$ can be covered by its $k$ proper subalgebras $\mathcal{A}_0,\ldots,\mathcal{A}_{k-1}$, where $k\in\mathbb{N}\setminus\{0,1,2,4\}$, such that $\mathcal{A}_i\not\subseteq\bigcup_{j\neq i}\mathcal{A}_j$ for every $i<k$. However, I am not sure if $k=4$ should really be excluded here. Thus, my question is the following:
Q: Let $\mathcal{A}$ be an infinite Boolean algebra. Do there always exist its proper subalgebras $\mathcal{A}_0,\mathcal{A}_1,\mathcal{A}_2,\mathcal{A}_3$ such that $\mathcal{A}=\bigcup_{i<4}\mathcal{A}_i$ and $\mathcal{A}_i\not\subseteq\bigcup_{j\neq i}\mathcal{A}_j$ for every $i<4$?
I suspect that a careful (and tedious) analysis of cases could show that the answer is negative (likewise for $k=2$), but perhaps there is some clever proof (or refutal) of it. I am also asking you about possible references to papers or books in which such problems are studied.
(I asked the same question at the Math Stack Exchange, but the interest was literally null.)
 A: It's a good idea for such questions to think about the finite case, since it indeed allows to solve the general case.

Proposition: every unital Boolean algebra $A$ of (possibly infinite) cardinal $\ge 16$ (i.e., whose spectrum has cardinal $\ge 4$) is a non-redundant union of 4 unital subalgebras.

Proof: by pull-back, it's enough to suppose that $A=\mathbf{Z}/2\mathbf{Z}^4$. Its unital subalgebras are the $A_R$, where $R$ ranges over equivalence relations on $\{1,2,3,4\}$, and $A_R$ is the set of quadruples that are constant on $R$-equivalence classes.
Define $A_{ij}=\{a\in A:a_i=a_j\}$ and $A_{ijk}=\{a\in A:a_i=a_j=a_k\}$.
We consider the four unital subalgebras $A_{12}$, $A_{13}$, $A_{14}$, $A_{234}$. If we forget the first, the three others do not cover $A$, since $(0,0,1,1)\notin A_{13}\cup A_{14}\cup A_{234}$; similarly for the second and third; also forgetting the fourth one does not give a cover since $(0,1,1,1)\notin A_{12}\cup A_{13}\cup A_{14}$. Still the four form a cover: indeed the only elements not in $A_{12}\cup A_{13}\cup A_{14}$ are $(0,1,1,1)$ and $(1,0,0,0)$ and they belong to $A_{234}$. This concludes that $A=\mathbf{Z}/2\mathbf{Z}^4$ is the non-redundant union of $A_{12}$, $A_{13}$, $A_{14}$, $A_{234}$.
If take for granted your claim for other values [edit: justified by Emil Jeřábek as a comment below, by generalizing the above argument], this yields:

Corollary: for every $k\in\mathbf{N}\smallsetminus\{0,2\}$, every infinite unital Boolean algebra can be written as non-redundant union of $k$ unital subalgebras.

(No reason to exclude 1 by restricting to proper subalgebras: assuming non-redundancy assumption excludes the whole subalgebra for $k>1$.)

Addendum: a classical lemma of B-H. Neumann (for groups) implies that whenever a Boolean algebra $A$ is written as non-redundant union of $k$ subalgebras ($k<\infty$), then each of those subalgebras has finite index in $(A,+)$. 
Also, in a unital Boolean algebra $A$, every subalgebra of finite index contains an ideal of finite index (with some explicit bound). And hence is obtained by pullback from a subalgebra in a finite quotient Boolean algebra.
