$\pi_{1}({\mathbb R}^{2} - {\mathbb Q}^{2})$ is uncountable

In case anyone doesn't like the arguments from uncountability, here's a more concrete argument:

Fix a base-point $(p,p)$ for an arbitrary $p\in\mathbb{R}-\mathbb{Q}$. For any $q\in\mathbb{R}-\mathbb{Q}$, $q\not=p$, consider the loop $L_q$ consisting of the following line segments: $(p,p)\rightarrow(p,q)\rightarrow(q,q)\rightarrow(q,p)\rightarrow(p,p)$

We claim that for any $q_1<q_2$, the loops $L_{q_1}$ and $L_{q_2}$ are non-homotopic. To see this, pick a rational $r$ such that $q_1<r<q_2$. Embed $\mathbb{R}^2-\mathbb{Q}^2$ into $\mathbb{R}^2-(r,r)$ (aka, the punctured plane). Clearly, $L_{q_1}\equiv 0$ while $L_{q_2}\equiv 1$, so the loops are non-homotopic, as desired.

So now that Jacob has explained why the space is path connected let us finish the problem. In his explanation he points out that there are an uncountable number of lines passing through $p$ and an uncountable number of lines passing through $q$ and since we are removing only countably many points not all of these lines can hit a point in $\mathbb{Q}^2$. So we can construct a path from $p$ to $q$ missing $\mathbb{Q}^2$, in fact more than one path. That gives us a loop, inside this loop will be an element of $\mathbb{Q}^2$ just by looking at the coordinates of $p$ and $q$. This shows that the loop is not contractible, ie it is a nontrivial element of $\pi_1(\mathbb{R}^2-\mathbb{Q}^2)$. Now you can proceed to construct more loops passing through $p$ and $q$ that are not contractible or homotopic to the previously constructed loop by picking different line segments as above. Now just construct another path from $p$ to $q$ and notice that an element of $\mathbb{Q}^2$ is inside each of your new loops (there should be at least 2 new loops, try drawing a picture).

Does this help?

Edit:

Here is, in my mind, a clearer construction of uncountably many non-homotopic loops. Let $p \in \mathbb{R}^2-\mathbb{Q}^2$ and $L$ a line in $\mathbb{R}^2-\mathbb{Q}^2$ not containing $p$. As established, there is a continuum of lines passing through $p$, and once we remove all of the ones passing through a point in $\mathbb{Q}^2$ we still have uncountably many. All remaining lines, with one possible exception, intersect $L$ at distinct points. We are left with a triangle for each pair of distinct remaining lines passing through $p$. There is a rational point inside each of these triangles (maybe this is something that was unclear, let me know and I think I can make this part more explicit). There is no way to move "homotope" the triangle past one of the rational points so that it would be homotopic to any adjacent triangle. Suppose two "loops" created in this fashion were homotopic, and suppose further that the differ in the second line passing through $p$, then there would have to be a homotopy between two paths which enclose a point of $\mathbb{Q}^2$ which is impossible.

If you want I can draw a picture on my tablet and try and post. (I don't really know the proper way to go about posting such a picture.)

The way you get uncountably many loops is similar to the way you get uncountably many for the Hawaiian Earrings. Choose a sequence of loops $\alpha_i$ whose diameters approach $0$. Then consider the infinite concatenations $\alpha_1^{\epsilon_1}*\alpha_2^{\epsilon_2}\cdots*\alpha_n^{\epsilon_n}*\cdots$ where $\epsilon_n=\pm 1$. Since the diameters go to zero, this really represents a loop in the space. However, there are clearly uncountably many such loops. The tricky part is showing that no pair of them is homotopic.