$\mathbb{R}^2 \setminus \mathbb{Q}^2$ connected? To show that it is connected, I have assumed for contradiction that it is not. It can therefore be separated into two disjoint, proper clopen subsets $A \sqcup B.$ I have shown that it is not possible that $A$ is both closed and open, for if it is closed it either (1) has boundary points and contains all of them, or (2) it has no boundary points. I have shown that $A$ cannot have a boundary point, for that would contradict the assumption that it is open (no neighborhood of a boundary point is fully contained within the set). So, $A$ has no boundary points. Now, I'm trying to show that $A$ cannot be proper if it has no boundary points. HERE is my difficulty. I know $\mathbb{R}^2$ doesn't "have" infimums and supremums, but is there a way I can use the infimums and supremums along each dimension to show that it is bounded? That is, if $A \subset \mathbb{R}^2$ is bounded along each of its dimensions, then it itself is bounded. However, I am having trouble formalizing this part. My argument comes across clunky. Is there a simple way to do this, or will a different direction prove better?
 A: Expanding on Wojowu's comment, it is far easier to show that $\mathbb{R}^{2}$ less any countable set is path connected. From this connectedness is of course implied.
The argument ultimately comes down to the fact that the set of lines in the plane containing a particular point is uncountable. 
Let $A\subseteq\mathbb{R}^2$ be a countable set and let $x,y\in\mathbb{R}^{2}\setminus A$. Let $\mathcal{L}_{x}$ be all lines in $\mathbb{R}^{2}$ that contain $x$. Then define $f:A\rightarrow\mathcal{L}_{x}$ by defining $f(z)$ to be the unique line $L\in\mathcal{L}_{x}$ that contains $z$. Then, because $A$ is countable and $\mathcal{L}_{x}$ is uncountable we have that $\mathcal{L}_{x}\setminus f(A)$ is uncountable. We can similarly define a set $\mathcal{L}_{y}$ of lines containing $y$ and a function $g:A\rightarrow\mathcal{L}_{y}$. Because $\mathcal{L}_{x}\setminus f(A)$ and $\mathcal{L}_{y}\setminus g(A)$ are uncountable we can choose nonparallel lines $L_{1}\in\mathcal{L}_{x}\setminus f(A)$ and $L_{2}\in\mathcal{L}_{y}\setminus g(A)$ which necessarily intersect. A path from $x$ to $y$ (or $y$ to $x$) in $\mathbb{R}^{2}\setminus A$ can be constructed using these lines.
A: For feedback:
Consider any two points $p = (x_1,y_1)$ and $q = (x_2,y_2).$
It is the case that $x_1 \in \mathbb{I}$ is irrational or $y_1 \in \mathbb{I}$ is irrational, and that $x_2 \in \mathbb{I}$ is irrational or $y_1 \in \mathbb{I}$ is irrational. Suppose without loss of generality that $x_1$ is irrational. I will now consider the two cases: the case in which $x_2$ is irrational and the case in which $y_2$ is irrational. Suppose $x_2$ is irrational. Then, there exists the line $\mathcal{L_1} \in \mathbb{R}^2 \setminus \mathbb{Q}^2$  connecting $(x_1,y_1)$ to $(x_1,i_1),$ where $i_1 \in \mathbb{I}.$ Moreover, there exists the line $\mathcal{L}_2 \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ connecting $(x_1,i_1)$ to $(x_2,i_1),$ where $i_2 \in \mathbb{I}.$ And, there exists a line $\mathcal{L}_3 \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ connecting $(x_2,i_2)$ to $(x_2,y_2).$ So, the path $P = \mathcal{L}_1 \cup \mathcal{L}_2 \cup \mathcal{L}_3 \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ is a path connecting $(x_1,y_1)$ to $(x_2,y_2).$ The proof is almost identical when considering the case that $y_2$ is irrational.
Is a proof like this even acceptable?
A: I was going to suggest a more direct argument.

Claim 1: Let $(a,b') \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ (which we will write as $X$) s.t. $b'$ is irrational. Then for any $a'$ irrational there is a path $P$ from $(a,b')$ to $(a',b')$ in $X$.

Indeed, let $P$ be the path $(\lambda a' + (1-\lambda) a, b')$; $\lambda \in [0,1]$. As $b'$ is irrational $P$ is in $X$.
Similarly, the following holds:

Claim 2: Let $(a',b) \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ (which we will write as $X$) s.t. $a'$ is irrational. Then for any $b'$ irrational there is a path $P$ from $(a',b)$ to $(a',b')$ in $X$.

Finally, we note the following: 

Claim 3: Let $(c',d'')$ and $(a',b')$ be s.t. $a',b',c'$ and $d'$ are all irrational. Then there is a path in $X$ from $(a',b')$ to $(c',d')$. 

Indeed, let $P_1$ be the path $(c'\lambda + a'(1-\lambda), b')$; $\lambda \in [0,1]$, and let $P_2$ be the path $(c', d' \tau + b'(1-\tau))$; $\tau \in [0,1]$; concatonate $P_1$ and $P_2$
So now let $(a,b)$ and $(c,d)$ be points in $X$. Use Claims 1 and/or 2 to find $(a',b')$ and $(c',d')$; $a',b',c',d'$ all irrational, such that there is a path from $(a,b)$ to $(a',b')$ in $X$, and from $(c,d)$ to $(c',d')$ in $X$. Then finish using Claim 3.
