Let $\gamma: \mathbb{R} \to \mathbb{R}^2$, $\gamma$ continuously differentiable with $\gamma'(t) \neq 0$, $\forall t \in \mathbb{R}$. Then, for all $t_1 \in \mathbb{R}$, there is a $\epsilon>0$ such that does not exists open set $A \subset \mathbb{R}^2$, with $ A \subset$ $\gamma(t_1 - \epsilon, t_1+ \epsilon)$.
My attempt:
For assumptions I guess I need to use the Inverse Function Theorem, so fixing $t_1 \in \mathbb{R}$ There is neighborhood $U$ of $t_1$ and a neighborhood $V$ of $\gamma(t_1)$ such that $\gamma: U \to V $ is a bijection and it's inverse $\gamma^{-1}:V \to U$ is differentiable.
Ok, $\gamma(U)$ is a open set, so I need to build a set $B$ inside of $\gamma(U)$ such that there is not open set inside of $B$,maybe Implicit function theorem? I dunno.
Can you help me?