As the question is asked, the answer is "no": if a continuous curve $\gamma:\mathbb{R}_{\geq 0}\to [0,1)^2$ is self-avoiding, i.e., injective, then the image $\gamma(\mathbb{R}_{\geq 0})$ is nowhere dense in $[0,1)^2$. Indeed, the images $\gamma([0,T])$ are compact, hence closed. Also, compactness implies the inverse map $\gamma^{-1}:\gamma([0,T])\to[0,T]$ is continuous, i.e., the image $\gamma([0,T])$ is homeomorphic to an interval. Hence, it cannot contain any balls, for if it did, removing a point from it would make the fundamental group non-trivial, while removing a point from an interval cannot do so. Baire's theorem now implies that $\gamma(\mathbb{R}_{\geq 0})=\cup_{T\in\mathbb{N}}\gamma([0,T])$ is nowhere dense.