Why C^1 class mapping with nonzero derivative not fill a square

Why C^1 class mapping with nonzero derivative not fill a square

Let $f: [0,1] \rightarrow \mathbb R^2$ be of class $C^1$ with $f'(t)\neq
(0,0)$. Why $f$ is not Peano-type curve, i.e. $f(I) \neq I\times I$.