This is probably a silly question but I want to understand why regular curves are 1-dimensional manifolds, something mentioned in several books. I assume they mean the trace of a curve, so let $\gamma\colon I \to \mathbb{R}^2$ be a regular curve, i.e. $C^{\infty}$ and $|\gamma'(t)|\neq 0$ for all $t\in I$. The trace of the curve $\gamma(I)\subseteq \mathbb{R}^2$ we endow it with the subspace topology of $\mathbb{R}^2$, thus it will be Hausdorff and second countable. My problem is with the property of being locally Euclidean. A regular curve is allowed to have self intersections, for example, the curve $\gamma\colon \mathbb{R}\to\mathbb{R}^2$ given by $\gamma(t)=(t^{3}-4t,t^2-4)$, has the following graph
So for example we would have to find an open set that is of the form an open ball $B(a,r)\subseteq \mathbb{R}^2$ intersect with $\gamma(I)$, containing $(0,0)$, such that there is a homeomorphism to $\mathbb{R}$. We would have something like a curved $X$ around $(0,0)$. How is that homeomorphic to $\mathbb{R}$? Intuitively I understand that around each point on the trace (not one of this singular points) we can find an open set in that topology that is homeomorphic to $\mathbb{R}$. I would appreciate your help to clarify this doubt I have.