Every countable subset $A\subset \mathbb{R}^n$, with $n\gt 1$, and indeed, every subset of size less than the continuum, has a path-connected complement. This is because for any two points in the complement, there is a foliation of continuum many paths joining them, and so one most of these paths lies lie entirely in the complement of $A$.
The argument in your final paragraph appears to be conflating conflate $\mathbb{R}^n-B^n$ with $(\mathbb{R}-B)^n$.(\mathbb{R}-B)^n$, but these are not generally the same and they cannot be equal when$B$is countable. 2 added 120 characters in body; deleted 11 characters in body Every countable subset$A\subset \mathbb{R}^n$, with$n\gt 1$, and indeed, every subset of size less than the continuum, has a path-connected complement. This is because for any two points in the complement, there is a foliation of continuum many paths joining them, and so one of these paths lies entirely in the complement of$A$. The argument in your final paragraph appears to be conflating$\mathbb{R}^n-B^n$with$(\mathbb{R}-B)^n$. 1 Every countable subset$A\subset \mathbb{R}^n$, with$n\gt 1$, and indeed, every subset of size less than the continuum, has a path-connected complement. This is because for any two points in the complement, there is a foliation of continuum many paths joining them, and so one of these paths lies entirely in the complement of$A\$.