Topological proof to show a complement set of a set is polygonally connected I started learning topology recently, so I'm very new to this.  
The question is as follows:

Given the set $A = \{ (x,y)\;;\;x, y\in\mathbb{Q} \}=\mathbb{Q}^2$, show that the complement set $A^c=\mathbb{R}^2\setminus A$ is polygonally connected.

As an attempt on this question, I plan to show that the set $A^c$ is connected, thus using a theorem, I can say $A^c$ is polygonally connected.
But here is the part that I got confused:
By definition, the set $A^c$ is connected iff it cannot be written as the union of $2$ nonempty separated sets.  
In this case, I believe $A^c$ is the set of all $(x,y)$ where $x,y$ belong to $\mathbb R-  \mathbb Q$.  But how can I express the fact that $A^c$ is not the union of $2$ nonempty separated sets ? 
Based on my knowledge, I know both sets $\mathbb Q$  and $\mathbb R-\mathbb Q$ are dense, but how is this helpful to solving this problem?  I think a reason is because for any collection of $(x,y)$ where $x,y $are in $\mathbb Q$, I can always find a point in $\mathbb R-  \mathbb Q$ which is a boundary point of that certain collection.  But is my thought ok ?
Would someone please help me on this?  I'm kinda lost >_<
Thanks ^^
 A: Pick two points in $(x_1,x_2), (y_1,y_2) \in A^c$.
At least one of $x_1,x_2$ and at least one of $y_1,y_2$ are irrational.
Suppose $x_1, y_1$ are irrational. Then $(x_1,s), (y_1,t) \in A^c$ for all $s,t \in \mathbb{R}$. Then the path $(x_1,x_2) \to (x_1,\sqrt{2}) \to (y_1,\sqrt{2}) \to (y_1,y_2)$ lies entirely in $A^c$.
Suppose $x_1,y_2$ are irrational. Then the path $(x_1,x_2) \to (x_1,y_2) \to (y_1, y_2)$ lies entirely in $A^c$.
The other combinations follow the same pattern.
A: Consider two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ in $A^c$.  For each point, at least one of the coordinates is irrational.
Say, for example, that $y_1 \in \mathbb{R} \setminus \mathbb{Q}$.  Then, you can travel horizontally along the line $y = y_1$, all the while, staying within $A^c$.  Similarly, if $x_1 \in \mathbb{R} \setminus \mathbb{Q}$, then the vertical line $x = x_1$ lies entirely in $A^c$.
A polygonal path connecting $P_1$ to $P_2$ can be found entirely in $A^c$ by zig-zagging horizontally and vertically, all the while maintaining at least one irrational coordinate.
