Why isn't $\mathbb{R}^2$ a countable union of ranges of curves? I came across this question on the topology board at AoPS, where it hadn't really received an answer. It seems interesting, but I'm not sure how to solve it. Hopefully an answer will be found here.
The question, from Dudley's Real Analysis and Probability, goes as

A $C^1$ curve is a function $t\mapsto (f(t),g(t))$ from $\mathbb{R}$ into $\mathbb{R}^2$ where the derivatives $f'(t)$ and $g'(t)$ exist and are continuous for all $t$. Show that $\mathbb{R}^2$ is not a countable union of ranges of $C^1$ curves.

In the google book there's also a hint to show the range of a $C^1$ curve on a finite interval is nowhere dense. 
So out of curiosity, what's a proper way to prove the hint that the range of a $C^1$ curve on a finite interval is nowhere dense? I've read through the first few sections of the book up to the point of the exercise, and it covers compactness, product topologies, complete and compact metric spaces and metrics on function spaces, and the theorems of Dini, Arzela-Ascoli, and the Stone Weierstrass Theorem. I'm putting a bounty to see if there's a basic solution using only this knowledge, even if it may not be the most elegant.

Added: I did a little research based on the comments I received. I know that $\mathbb{R}^2$ is a complete metric space, and the Baire Category Theorem says every complete metric space is of the second category. I can decompose a curve into countably many curves defined on finite intervals, (like intervals of length $1$ maybe?). Then if $\mathbb{R}^2$ is the countable union of ranges of curves, it would also be the countable union of the ranges of these countably many smaller curves with finite domains. However, if each curve on a finite interval has nowhere dense range, then $\mathbb{R}^2$ would be the union of countable many nowhere dense sets and then be of the first category, contrary to the Baire Category Theorem.
So why exactly is the range of a curve on a finite interval nowhere dense? It seems intuitively true. If I let $B=\{(f(t),g(t))\mid t\in I\}$ for some finite interval $I\subset\mathbb{R}$ be the range of some $C^1$ curve, then my feeling is this range looks like some "segment" in $\mathbb{R}^2$. If $U\subset\mathbb{R}^2$ is any open set, I want to find an open $V\subset U$ where $B\cap V=\emptyset$. Intuitively, it seems like I could just take some open ball of small enough radius in $U$ that doesn't meet $B$ in the plane, so the range is nowhere dense. Is there a more formal way to express this? Thanks.
 A: A $C^1$ curve on a compact interval has nowhere dense range: it is because it is Lipschitz, since its derivative is bounded. Lipschitz functions cannot increase the Hausdorff dimension, which is less than $2$ for an interval. So the range has two dimensional measure $0$, and it is closed, so it contains no open set.
A: You can construct a tubular neighborhood of such a curve having any desired area.  It is the locus of points 
$$(f(t),g(t)) + u \frac{\varepsilon(t)}{2}(g'(t),-f'(t))$$
where $t$ ranges over $\mathbb{R}$, $u$ ranges from $-1$ to $1$, and $\varepsilon(t)$ is suitably chosen.   For instance, for all $t$ where the derivative of the curve is nonzero, let $\varepsilon(t) = |\varepsilon|/(\pi(1+t^2)||(f'(t),g'(t))||)$ (and otherwise let $\varepsilon(t)=0$) to drive the area down to $\varepsilon +  O(\varepsilon^2)$, $\varepsilon \gt 0$.  (Simple estimates will make this rigorous: it doesn't take any measure theory or fancy analysis.  Just break the integral down into intervals over which neither $f'(t)$ nor $g'(t)$ varies much.)
Given a countable sequence of curves $(\gamma_i)_{1 \le i}$, create such tubular neighborhoods with $\varepsilon_i(t)$ chosen to make the tubular neighborhood of $\gamma_i$ less than $2^{-i}$ in area.  The area of the union of those neighborhoods--which obviously includes the union of the curves--cannot exceed $\sum_i \varepsilon_i = 1$, whence it cannot fill $\mathbb{R}^2$.
This approach generalizes to differentiable submanifolds of $\mathbb{R}^n$.
A: The image of a $C^1$ curve has measure zero by Sard's lemma, and the countable union of sets of measure zero has measure zero. 
A: Here's an answer avoiding measure theory, or any advanced methods. First, if a curve $\gamma\colon\mathbb{R}\to\mathbb{R}^2$ is $C^1$ then, for any $a < b$, the restriction of $\gamma$ to $[a,b]$ has finite length $\int_a^b\vert\gamma^\prime(t)\vert\,dt$.
Now, a finite length curve cannot be dense the unit square $[0,1]^2$. For any positive integer $n$, consider the $(n+1)^2$ points $(i/n,j/n)$, $0\le i,j\le n$. The distance between any two of them is at least $1/n$. So, any curve which is dense in the unit square must join these points, and hence have length at least $((n+1)^2-1)/n=n+2$. Let $n$ go to infinity. By scaling, this shows that a finite length curve cannot be dense in any open subset of $\mathbb{R}^2$.
Now suppose that we have a countable set of $C^1$ curves $\gamma^i\colon\mathbb{R}\to\mathbb{R}^2$ ($i=1,2,\ldots$). Then, $\gamma^{ij}=\gamma^i\vert_{[-j,j]}$ ($i,j=1,2,\ldots$) is a countable set of curves of finite length, so their images are compact and nowhere dense subsets of $\mathbb{R}^2$.
Finally, the Baire category theorem says that a countable sequence of closed nowhere dense subsets of $\mathbb{R}^2$ cannot cover $\mathbb{R}^2$.
