Infinitely countable subset of $\mathbb{R}^2$ is connected. Let $A\subset \mathbb{R}^2$ be an infinite countable subspace. 
Can $A$ be connected?
I have seen plenty of proofs that show that $\mathbb{R}^2\setminus A$ is connected and I understand those.  However, none of them talk about whether $A$ is connected.  
I would think it would be.  By the same logic as those other proofs here is my proof:
Let $x,y\in A$.  Now there are uncountably many paths from $x$ to $y$ in the plane.  Since $A$ is countable then there exists at least one path from x to y that would start at x, leave $A$, and come back to $A$ and end at $y$. (Paths will leave $A$)  However, I don't see how that would stop there from being a path from x to y in A. 
There was another theorem that said let $X$ be Hausdorff and $A\subset X$ with $A$ connected, then $A$ contains 1 or infinitely many elements.  
Now, I tried using this theorem (because $\mathbb{R}^2$ is Hausdorff), but my logic started to confuse me. 
If $A$ is not connected, then I was going to use this theorem in my proof of contradiction.  So suppose $A$ is connected, then this theorem says $A$ has one or infinitely many points, but $A$ has infinitely many elements.  So, we have no contradiction.  
And if we suppose $A$ is not connected, then we can't use this theorem because the assumptions are not met. 
This Theorem might not even help us at all.
But please help me! 
 A: Any countable subset of $\mathbb{R}^2$ (with at least two elements :P) is indeed disconnected; but I think it takes a different argument than what you sketch to show this.
Here's a proof outline. Let $A\subseteq\mathbb{R}^2$ be countable and fix distinct $a, a'\in A$. Say that a positive real $r$ is good if 


*

*$r<d(a, a')$, and

*for no $b\in A$ do we have $d(a, b)=r$.

Exercise: Since $\mathbb{R}$ is uncountable, there is some good $r$.

This will use the fact that $d(a, a')>0$ - so in particular there are uncountably many positive reals less than $d(a, a')$.
OK, let $r>0$ be good. Now consider $B_r(a)$, the ball of radius $r$ centered on $a$.

Exercise: For each $b\in A\setminus B_r(a)$, there is some open ball $U_b\ni b$ with $U_b\cap B_r(a)=\emptyset$.

This is where the goodness of $r$ is used.
OK, now consider the two sets $$B_r(a)\quad\mbox{ and }\quad \bigcup_{b\in A\setminus B_r(a)} U_b.$$

Exercise: these are nonempty open sets which are disjoint and cover $A$.

Note that $a'\in \bigcup_{b\in A\setminus B_r(a)} U_b$.
So $A$ is not connected.

"Hang on!" you might say, "This is a topology problem. Why drag a metric into things?" 
There is indeed a way to write this argument without invoking a metric at all. Fix distinct $a, a'\in A$, $A\subseteq\mathbb{R}^2$ countable and nonempty. Then say that an open set $U$ is good if


*

*$U$ contains $a$ but not $a'$, and

*For every $b\in A$, either $b\in U$ or $b\not\in \overline{U}$ (where "$\overline{U}$" denotes the closure of $U$). 
Then, we only need to show

There is an uncountable family $U_i$ of open sets containing $a$ but not $a'$, which have disjoint boundaries.

Then a counting argument shows that there is a good open set, and the rest of the argument goes through unchanged.
Note, though, that the highlighted step above is easiest to do if we have the metric description of $\mathbb{R}^2$ at our disposal.
