Topologist's Sine Curve not a regular submanifold of $\mathbb{R^2}$?

Let $X$ denote the $x$-axis. Assume that $U\ni p$ is open and $\phi(U)$ is an open subset of $\mathbb R^2$ with $\phi:U\to \phi(U)$ a homeomorphism. If $\phi(U)$ intersects $X$ in several components, let $C$ denote the connected component containing $\phi(p)$. Since $\phi(U)$ is open, $\phi(U)\cap X$ is open in $X$. But $X\approx\mathbb R$ is locally connected, therefore components of open subsets are open. So $C$ is open and $X-C$ is closed in $X$. Since $X$ is closed in $\mathbb R^2$, $X-C$ is closed in $\mathbb R^2$ and $\phi(U)-(X-C)$ is open.
Applying $\phi$ we get a homeomorphism $\phi^{-1}(\phi(U)-(X-C))\approx \phi(U)-(X-C)$, the later of which intersects the $x$-axis in just a single component, which is a contradiction.