Let $\phi : [0, 1] \to [0, 1]$ be the Cantor-Lebesgue function, and $\alpha : [0, 1] \to \Bbb{R}^n$ be a space-filling curve.
Since $\phi$ is stationary outside the Cantor set, it is locally constant almost everywhere. That is, $\beta := \alpha \circ \phi$ is also locally constant everywhere, allowing it to be differentiable a.e.. (In fact, $\beta' = 0$ a.e.!) On the other hand, $\phi$ is continuous and surjective. Thus $\beta$ is also a continuous path and the image of $\beta$ coincides with $\alpha$. That is, $\beta$ is also a space-filling curve. Therefore this serves as a counter-example.
Let $f : [0, 1] \to \Bbb{R}^n$ be a differentiable curve in $\Bbb{R}^n$ ($n \geq 2$). Let $\Gamma = f([0, 1])$ be the image of $f$ in $\Bbb{R}^n$.
Assuming that $|f'|$ is Lebesgue integrable, we can prove the first statement.
Theorem. [7.21, Rudin] If $f : [a, b] \to \Bbb{R}$ is everywhere differentiable and its derivative $f'$ is Lebesgue integrable, then
$$ f(b) - f(a) = \int_{a}^{b} f'(t) \; dt. $$
This theorem immediately implies the following corollary:
Corollary. Let $f : [0, 1] \to \Bbb{R}^n$ be everywhere differentiable and its derivative $f'$ is Lebesgue integrable. Then for any $\epsilon > 0$, then there exists a partition $\Pi = \{ 0 = x_0 < \cdots < x_N = 1 \}$ such that for
$$\epsilon_k = \sup \{|f(x) - f(y)| : x, y \in [x_{k-1}, x_k] \}, \quad (1 \leq k \leq N)$$
we have
$$\epsilon_k \leq \epsilon \quad \text{and} \quad \sum_{k=1}^{N} \epsilon_k \leq \| f' \|_1.$$
Proof. Since $f'$ is Lebesgue integrable, it is absolutely continuous. Thus there exists $\delta > 0$ such that whenever a measurable subset $E \subset [0, 1]$ satisfies $|E| < \delta$, we have $\int_E |f'| < \epsilon$. Now let $\Pi = \{x_k\}$ be a partition of $[0, 1]$ into subintervals with length less than $\delta$. Then for each $x_{k-1} \leq x < y \leq x_k$,
$$ |f(y) - f(x)| = \left| \int_{x}^{y} f' \right| \leq \int_{x}^{y} |f'| \leq \int_{x_{k-1}}^{x_k} |f'| < \epsilon. $$
Thus
$$ \epsilon_k \leq \int_{x_{k-1}}^{x_k} |f'| < \epsilon $$
and the conclusion follows. ////
Remark. If $f'$ is bounded, then it is Lebesgue integrable. Also, in this case, the conclusion of the Corollary follows directly by mean value theorem.
Now let $\epsilon > 0$ and $\Pi$ be a corresponding partition of $[0, 1]$ by Corollary. Then we can cover $\Gamma$ by balls $B_{2\epsilon_k}(f(x_k))$ for $k = 1, \cdots, N$. Thus
$$ |\Gamma| \leq \sum_{k=1}^{N} \left|B_{2\epsilon_k}(f(x_k))\right| \leq \sum_{k=1}^{N} C_n \epsilon_k^n \leq C_n \epsilon^{n-1} \sum_{k=1}^{N} \epsilon_k \leq C_n \epsilon^{n-1} \| f' \|_1, $$
where $C_n$ is a constant depending only on the dimension $n$. (In fact, we can choose $C_n = |B_2|$.) Thus taking $\epsilon \to 0$, we have the desired result.
Best Answer
Basically, just do what the hint says. The idea is to cover the curve (i.e. the graph of your function $f:\mathbb R \to\mathbb R$) by rectangles such that the sum of their areas (measures) is smaller than $\epsilon>0$ and do so for every $\epsilon>0$. This would show that $m(C)<\epsilon$ for every $\epsilon>0$ and thus $m(C)=0$.
Let's do this.
Lemma. Let $g:[a,b]\to\mathbb R$ be a continuous function and let $\Gamma_g = \{(x,g(x))|\;x\in[a,b]\}$ be the graph of $g$. Then $m(\Gamma_g)=0$.
Proof. The function $g$ is uniformly continuous, since $[a,b]$ is compact. Let $\epsilon>0$. Then there exists a $\delta>0$, such that for each pair of points $x,y\in [a,b]$ such that $|x-y|<\delta$, the inequality $|g(x)-g(y)|<\epsilon$ holds. We may assume without loss of generality that $\delta<b-a$ (since if some $\delta\geq b-a$ works for this choice of $\epsilon$, any other number $0<\delta_1<b-a$ would also work.)
Let $n\in\mathbb N$ be the smallest natural number such that $n\delta> b-a$. Then there exist numbers $a=x_0<x_1<\ldots<x_n=b$ on the interval $[a,b]$, such that for each $i=1,2,\ldots,n$ we have $|x_i-x_{i-1}|<\delta$. (Proving this is an easy exercise.) For each $i=1,2,\ldots,n$ let $y_i=g(x_i)$.
Now, for each $i=1,2,\ldots,n$ and each point $z\in[x_{i-1},x_i]$, we have $g(z)\in[y_i-\epsilon,y_i+\epsilon]$ because for such $z$ the inequality $|z-x_i|<\delta$ holds, and thus by definition of $\delta$, we also have the inequality $|g(z)-y_i|<\epsilon$. This implies that $\Gamma_g\subseteq\bigcup_{i=1}^n[x_{i-1},x_i]\times[y_i-\epsilon,y_i+\epsilon]$ and therefore $$m(\Gamma_g)\leq\sum_{i=1}^n m([x_{i-1},x_i]\times[y_i-\epsilon,y_i+\epsilon])\leq\sum_{i=1}^n 2\delta\epsilon = 2\delta\epsilon n$$ Since $n$ is the smallest natural number for which $n\delta>b-a$, we must also have $b-a+\delta\geq n\delta$ (since otherwise we would have $b-a+\delta<n\delta$ which would imply $b-a<(n-1)\delta$, which would contradict our choice of $n$). This means that $2\delta\epsilon n\leq 2\epsilon(b-a+\delta)<4\epsilon(b-a)$ since $\delta<b-a$.
So, we showed that for every $\epsilon>0$ the inequality $m(\Gamma_g)<4\epsilon (b-a)$ holds. This is only possible if $m(\Gamma_g)=0$ and thus the proof is complete. $\square$
Now, to show this for $f:\mathbb R\to\mathbb R$ notice that the graph of $f$ (defined by $\Gamma_f=\{(x,f(x))|\;x\in\mathbb R\}$) is the union of graphs of restrictions to $[-n,n]$. More precisely: define for each $n\in\mathbb N$ a function $f_n:[-n,n]\to\mathbb R$ by the formula $f_n(x) = f(x)$. Then $$\Gamma_f=\bigcup_{n=1}^\infty\Gamma_{f_n}.$$ Therefore, $$m(\Gamma_f)\leq\sum_{n=1}^\infty m(\Gamma_{f_n})=\sum_{n=1}^\infty 0 = 0$$ and thus $m(\Gamma_f)=0$. We are done.