Discrete subset of $\mathbb R^2$ such that $\mathbb R^2\setminus S$ is path connected. Let, $S\subset \mathbb R^2$ be defined by $$S=\left\{\left(m+\frac{1}{2^{|p|}},n+\frac{1}{2^{|q|}}\right):m,n,p,q\in \mathbb Z\right\}.$$ Then, which are correct?
(A) $S$ is a discrete set.
(B) $\mathbb R^2\setminus S$ is path connected.
I think $S$ is a discrete set. If we fix any three of $m,n,p,q$ then the set which we get is countable. Thus we get $S$ as the union of four countable sets. So $S$ is countable & so $S$ is discrete. But I am not sure about it..If I am wrong please detect my fallacy and give what happen? 
If $S$ is a discrete set then $\mathbb R^2\setminus S$ is path connected. But if NOT then what about the set $\mathbb R^2\setminus S$ ?
Edit : I know that a set $S$ is said to be discrete if it is closed and all points of it are isolated.
Am I correct ? Please explain.

Hint: Try to think of the graph of S near any of its limit points (m,n). It will somewhat look like kitchen sink filter which has more and more holes as you approach towards its centre (a limit point). 
