Let $O$ be an open set in a topological space $X$ and $p\in O,q\in X-\bar{O}$. Does every continuous curve from $p$ to $q$ pass trough $\partial O$? Suppose that $X$ is a topological space and $O\neq \emptyset$ is an open subset of $X$. Let $c:[0,1]\to X$ be a continuous curve from $p\in O$ to $q\in X\setminus\overline{O}$. Is it then true that $c$ passes trough $\partial O$, i.e. $c([0,1])\cap \partial O\neq \emptyset$?
My idea: We have $c(0)=p$ and $c(1)=q$. If $c(0.5)\in \partial O$, we are done. If $c(0.5)\in O$ or $c(0.5)\in  X\setminus \overline{O}$, we restrict ourselves to the interval $[0.5,1]$ or $[0,0.5]$ and continue. In this way, by using the Cantor intersection theorem, we can find $t_0\in [0,1]$ and sequences $x_n,y_n\in [0,1]$ with $x_n,y_n\to t_0$ such that $c(x_n)$ lies in $O$ and $c(y_n)$ lies in $X\setminus \overline{O}$.
Now let $M$ be any open set which contains $c(t_0)$. Then $c^{-1}(M)$ is an open set around $t_0$ . Since $x_n,y_n$ converge to $t_0$, it follows that $M$ contains points from $O$ and $X\setminus \overline{O}$ which shows that $c(t_0)$ is a boundary point of $O$.
This question is based on Lemma 6.2. on page 94 of Lee's "Riemannian Manifolds" (the part after (6.1); I have replaced the manifold $M$/the geodesic ball with an arbitrary topological space/open set, so the statement might not be correct.)
Lee argues that $X\setminus \partial O$ is disconnected so $c$ has to pass trough $\partial O$ since $p$ and $q$ lie in different components. However I don't really understand his argument.
 A: Let $c:[0,1]\to X$ be continuous with $c(0)=p$ and $c(0)=q$. Suppose, for a contradiction, that $c([0,1])\subseteq O \cup (X\setminus \bar{O})$. Set $A = c^{-1}(O)$ and $B=c^{-1}(X \setminus \bar{O})$. We have that $A,B$ is a partition of $[0,1]$ into non-empty open sets, contradicting the fact that $[0,1]$ is connected.
A: Your argument seems fine. I would consider the number $s=\sup\left\{t\in[0,1]\,\middle|\,c(t)\in O\right\}$. Since $0\in\left\{t\in[0,1]\,\middle|\,c(t)\in O\right\}$ and since $\left\{t\in[0,1]\,\middle|\,c(t)\in O\right\}\subset[0,1]$, the definition of $s$ makes sense. And it follows from the definition of $s$ that there are points $x\in[0,1]$ arbitrarily close to $s$ such that $c(x)\in O$. So, $c(x)\in\overline O$. But you cannot have $c(x)\in O$, because otherwise there would be a $\varepsilon>0$ such that $(s-\varepsilon,s+\varepsilon)\subset[0,1]$ and that $c\bigl((s-\varepsilon,s+\varepsilon)\bigr)\subset O$, and then$$\overbrace{c+\frac\varepsilon2}^{\phantom O\in O}>s=\sup\left\{t\in[0,1]\,\middle|\,c(t)\in O\right\}.$$So, $c(x)\in\overline O\setminus O=\partial O$.
