$R$ be an infinite Boolean ring ($a^2=a,\forall a \in R$) with unity , then must $R$ be uncountable? Let $R$ be an infinite Boolean ring ($a^2=a,\forall a \in R$) with unity , then is it true that $R$ must be uncountable ?
 A: No, an infinite Boolean ring with unity need not be uncountable.
The Löwenheim-Skolem theorem says that any first-order theory with an infinite model has models of all infinite cardinalities.
The notion of Boolean ring with unity is axiomatizable in first-order logic, and that first-order theory has infinite models (any power set of an infinite set, under symmetric difference and intersection).  So it has a countably infinite model.
I think this is a specific example: the collection of all sets of integers which are either finite or have finite complement, under symmetric difference and intersection.
A: An elementary example is not hard.
If $F_2$ is the ring with two elements, take the subalgebra of $\prod_{i=0}^\infty F_2$ generated by $\oplus_{i=0}^\infty F_2$ and the identity of the product ring.
There are obviously only countably many sequences of finite support in $\oplus_{i=0}^\infty F_2$, and the subalgebra generated is literally $\oplus_{i=0}^\infty F_2\cup \{1+x\mid x\in \oplus_{i=0}^\infty F_2\}$, so the whole thing is countable.

Later:
Just noticed the ring of sets described at the end of the other post. This ring is isomorphic to that one.
