Consider $\mathbb{D} \subseteq \mathbb{R}^2$ denote the unit disk. We call a function $\gamma : [0,1] \to \mathbb{D}$ that is continuous on $[0,1]$ and differentiable on $(0,1)$ a curve. Consider the following question:
Is it possible to cover $\mathbb{D}$ with countably many curves? That is, do there exist a countably family of curves $\{\gamma_n : n \in \mathbb{N}\}$ such that $\mathbb{D} \subseteq \bigcup_{n=0}^\infty\mathrm{ran}(\gamma_n)$?
This problem can be easily solved using measure theory: The (Lebesgue) measure of $\mathbb{D}$ is positive, while the measure of each curve is $0$, so $\bigcup_{n=0}^\infty \mathrm{ran}(\gamma_n)$ also has measure zero by $\sigma$-additivity. Hence such a family of curve does not exist. (Note also that as remarked by @freakish in his comment, the condition for $\gamma_n$'s being differentiable is necessary)
I would like to know if there is a proof of this using elementary analysis methods. As a reference, I wish to restrict the methods used in the proof to Chapter 1-3 of baby Rudin (and if possible, as little topology as possible).