Let $B$ a Boolean Algebra and $St(B)$ its stone space. I want to prove that $B$ is countable if and only if $St(B)$ is separable (i.e. there is a countable dense subset).
For the first implication: the base of colpen sets of the form $N(b)=\{U \in St(B): b\in U\}$ is countable. By AC I can choose $U_b \in N(b)$ and the set $\{U_b : b \in B\}$ is dense and countable. Is there a way not involving AC?
For the second implication, I have no idea.
You cannot show what you want: let $X$ be a separable compact zero-dimensional space that is not second countable (the double arrow will do, or $\{0,1\}^{\omega_1}$ e.g.), then its clopen algebra is a counterexample to your claim.
In fact $B$ is countable iff $\textrm{St}(B)$ is second countable, where the reverse is shown by refining the standard clopen base $N(b), b \in B$ to a countable one.