This is question 1.2.17 from Hatcher, Algebraic Topology:
Show that $\pi_1 ( \mathbb{R}^2 - \mathbb{Q}^2)$ is uncountable
My line of thinking is trying to show that there exists an injection $$ \begin{align} \rho :\mathbb{R} \setminus \{ 0 \} & \to \pi_1 (\mathbb{R}^2 - \mathbb{Q}^2, (\sqrt{2}, 0) )\\ r & \mapsto [I_r] \end{align} $$ where $I_r$ is the square loop that goes around the corners $$ (\sqrt{2}, 0) \to (\sqrt{2}, r) \to (-\sqrt{2}, r) \to (-\sqrt{2}, 0) \to (\sqrt{2}, 0) $$
Essentially, I'm trying to prove that if $a, b \in \mathbb{R}$ and $a \neq b$ then $[I_a] \neq [I_b]$. My intuition is that no point in $I_a$ is path connected to any point in $I_b$ so there's not going to be a homotopy between them; I'm unsure if this is enough and am looking for a rigourous way of expressing this.
Also, if anyone knows a nice, quick way of doing this question along a different line of reasoning, that would be appreciated!
Your idea is fine. However, for rational $r$, your loop would pass through $(0,r)\in\Bbb Q^2$. You can mend this by taking only irrational $r$ (these are also uncountably many).
To see that really $[I_a]\ne [I_b]$ for $a\ne b$, note that there exists a rational $q$ between $a$ and $b$ and that exactly one of the two loops is contractible in $\Bbb R^2\setminus\{0,q\}$. This says that the images of $[I_a]$ and $[I_b]$ under the canonical map $\pi_1(\Bbb R^2-\Bbb Q^2,(\sqrt 2,0))\to \pi_1(\Bbb R^2-\{(0,q)\},(\sqrt 2,0))$ differ.
Remark: Can you see why $\pi_1(\Bbb R^2-A)$ is also uncountable where $A=\{\,(\tfrac1n,0)\mid n\in\Bbb N\,\}$? One needs a slightly different argument (or maybe the same argument, but with more elaborate loops)