Why isn't there a continuously differentiable surjection from $I \to I \times I$? I was asked this question recently in an interview. Why can't there be a a continuously differentiable map $f \colon I \to I \times I$, which is also surjective? In contrast to just continuous, where we have examples of space filling curves.
I have one proof which is direct via Sard's theorem. But the statement seems simple enough to have a proof using at most implicit (and inverse) function theorem.

The image of any continuously differentiable function $f$ from $I$ to $I\times I$ has measure zero; in particular, it cannot be the whole square.
To see this, note that $\|f'\|$ has a maximum value $M$ on $I$. This implies that the image of any subinterval of $I$ of length $\epsilon$ lies inside a circle of diameter $M\epsilon$, by the mean value theorem. The union of these $\approx 1/\epsilon$ circles, which contains the image of $f$, has measure $\approx \pi M^2\epsilon$. Now let $\epsilon$ tend to $0$.
