$\textbf{Question}$ Suppose $f:[0,1]\rightarrow\mathbf{R}^{2}$ is a continuously differentiable, 1-1 function. If $f(a)\in f([0,1])$, then should there be some open ball $B(f(a),\epsilon)$, centered at $f(a)$ with small radius $\epsilon$ so that $f([0,1])\cap B(f(a),\epsilon)=f(a-\delta_{1},a+\delta_{2})\neq f([0,1])$ for some $\delta_{1},\delta_{2}>0$?
One observation was that if $f$ is a space-filling curve, then the answer is no. But the conditions on $f$ requires it rectifiable, and also because it's 1-1 , its image should be a nowhere dense closed set. So, $f([0,a-e])\cup f([a+e,1])$ for small $e>0$ is closed and its complement is nonempty open, so there is some open ball centered at $f(a)$ that doesn't intersect with $f([0,a-\epsilon])\cup f([a+\epsilon,1])$. So there is some $B(f(a),\epsilon)$ s.t. $f([0,1])\cap B(f(a),\epsilon)\subset f(a-\delta_{1},a+\delta_{2})$ for some $\delta_{1},\delta_{2}>0$. But should $\supset$ also hold for some suitable choices?
In the form stated the claim is false. You may consider e.g. $t\in [0,1]\mapsto (x,y)=(t^6, t^3 \sin \frac{1}{t})$. The envelope (where |sin| is one) is of the form $x=y^2$. The distance to the origin is a wildly oscillating function as $t\rightarrow 0$. You may also construct a $C^\infty$ counter-example.
It is true if we only ask that for any $\delta>0$ we have inclusion $f([0,1])\cap B(f(a),\epsilon) \subset f(a-\delta,a+\delta)$ for some $\epsilon>0$.
Assume the contrary and let $a\in[0,1]$ be a point where the condition fails. Then $f^{-1}B(f(a),1/n)$ does not shrink to $a$, i.e. there is $\delta>0$ and $x_n\in[0,1]$ so that $|x_n-a|\geq \delta$ and $f(x_n)\rightarrow f(a)$. Now $K=[0,1]\setminus ]a-\delta,a+\delta[$ is compact so there is a convergent subsequence $x_{n_k}\rightarrow b\in K$ and by continuity $f(a)=f(b)$ but $a<>b$.