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Let $f: R \rightarrow R^2$ be a $C^1$ function.

Show: for any interval $I\subset R$, $f(I)$ has zero Jordan measure.

Intuitively, I'm still not familiar with Jordan measure, beyond being able to prove that finite sets have measure zero by assigning each point some interval $[p, p+\epsilon/?]$, but I'm not sure intuitively how to extend this to curves

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This will only work if $I$ is closed and bounded, because as far as I know, Jordan measure is only defined for bounded subsets of the plane.

Without loss of generality, assume $I = [0,1]$. Let $|f'| \leq M$. Then the mean value theorem for vector-valued functions shows that $|f(x)-f(y)|\leq M|x-y|$.

Now let $n \geq 1$. For each $k = 0, \dots, n-1$ we have $$f([k/n,(k+1)/n]) \subseteq E_k,$$ where $E_k$ is the square $f(k/n) + [-M/n,M/n] \times [-M/n,M/n]$. Thus $f(I)$ can be covered by the union of $n$ squares each with area $4M^2/n^2$. This union is a simple set with area $\leq 4M^2/n$, which tends to zero as $n$ becomes large.

It follows that $f(I)$ has Jordan measure zero.

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    If $I$ is bounded but not closed, then $f(\overline I)$ still has zero Jordan content, and that holds for its subset $f(I)$ too. However if $I$ is unbounded, then the result is not necessarily true, even if $f(I)$ happens to be bounded. – 2017-02-17
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    @HenningMakholm True, given that $f$ is assumed defined on $R$. I suppose I was imagining that $f$ might be defined only on $I$. – 2017-02-17