Connectedness of the complement of small subsets (extended question) The following questions occurred to me while browsing this site and looking at Exercise 20 here. 
Question 1. Let $n>1$. Does there exist a countable dense subset $A\subset\mathbb{R}^n$ for which the complement $\mathbb{R}^n\setminus A$ is disconnected?
EDIT. I deleted an erroneous paragraph. Let me add two more questions, the first one being Gerald Edgar's comment here, the second one correcting the erroneous paragraph.
Question 2. Let $n>1$. Is it true that for any subset $A\subset\mathbb{R}^n$ of Hausdorff dimension less than $n-1$ the complement $\mathbb{R}^n\setminus A$ is connected?
It seems that Joel's argument answers this in the affirmative as well.
Question 3. Let $n>1$. Are there two countable dense subsets $A,B\subset\mathbb{R}^n$ whose complements are not homeomorphic?
 A: The answer to Question 3 is negative;
this is an immediate consequence of
the following classical theorem:


Theorem.   For all  $n\ge 1$, if $A$ and $B$ are countable dense subsets of $\Bbb{R}^n$, then there is a homomeomorphism $f: \Bbb{R}^n\rightarrow \Bbb{R}^n$ such that $f(A)=B$.


Historical note: In the above theorem, the $n=1$ case is due to Cantor; later and independently the general case was established by Fréchet [Les dimensions d’un ensemble abstrait, Math. Ann. 68 (1910), 145–168] and Brouwer [Some remarks on the coherence type $\eta$, Proc. Akad. Amsterdam
15 (1913), 1256–1263]. 
The above theorem also holds for the Hilbert cube sitting in for $\Bbb{R}^n$, as shown by M.K. Fort in his paper Homogeneity of infinite products of manifolds with boundary, Pacific J. Math. 12 (1962), 879–884.
A: 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 most of these paths lie entirely in the complement of $A$. 
This observation also appears to answer the exercise in your link.
The argument in your final paragraph appears to conflate $\mathbb{R}^n-B^n$ with $(\mathbb{R}-B)^n$, but these are not generally the same and they cannot be equal when $B$ is countable. 
