Suppose $\Omega\subset \mathbb{R}^n$ is open and bounded, and let $\partial\Omega$ the boundary of $\Omega$. Fix $x \in \overline{\Omega}=\Omega \cup \partial\Omega$ and let $y:[0,+\infty)\to \mathbb{R}^n$ be a continuous function such that $y(0)=x$. Now define $$B:=\{t\geq 0\,:\,y(t)\in \partial\Omega\}.$$ I have to prove that:
1)If $B=\emptyset$, then $y(t)\in \Omega$ for all $t\geq 0$.
2)If $B\neq \emptyset$ then, letting $\overline{t}:=\inf B$, one has $y(t)\in\Omega $ for all $0\leq t< \overline{t}$ and $\overline{t}\in B$.
I think I have to use continuity of $y$ and the definition of boundary of a set in order to prove it, but I'm a little bit confused on how to proceed.
Any suggestion would be really appreciated.
Consider $A=y^{-1}(\Omega),$ and $C=y^{-1}(\Bbb R^n-\bar\Omega),$ which are open because $y$ is continuous, and are disjoint. We see that $A,B,C$ partition the half-line $X=[0,\infty)$ (note $A$ does not meet $B$ because $\partial\Omega$ does not meet $\Omega$ because $\Omega$ is open so $\Omega^\circ=\Omega$). We also get $B=X-(A\cup C)$ so $B$ is closed.
For part (1), we observe that as $B=\varnothing$ and $X$ is connected, and $0\in A,$ we must have $C=\varnothing$ else we would have disconnected $X$ into two open sets, $A,C.$
For part (2), consider $[0,\bar t)=Y\subset X$ which is connected, and so $Y=X\cap Y=(Y\cap A)\sqcup (Y\cap B)\sqcup(Y\cap C)$. By the definition of $\bar t,$ we have that $B\cap Y=\varnothing$ so by the same argument as above $Y\subset A$ which gives the first part, and the because $B$ is closed, we get $\bar t\in B$.