Can anyone verify this proof, please?
f) if $x_0$ is an interior point and $x_1$ is an exterior point in relation to the set $M \subset \mathbb{R}^m$, then the support of any path with endpoints $x_0, x_1$ intersects the boundary of the set $M$.
Proof: Let $M \subset \mathbb{R}^m$, $x_0$ be an interior point of $M$ and $x_1$ be an exterior point of $M$. Let $\Gamma : [\alpha , \beta ] \to \mathbb{R}^m$ be any path joining $x_0$ and $x_1$, such that $\Gamma (\alpha ) = x_0$ and $\Gamma (\beta ) = x_1$. Let $I = \{\gamma \in [\alpha , \beta ] : \Gamma (\gamma ) \text{ is an interior point of } M\}$. Since $\Gamma (\beta ) = x_1$ is an exterior point of $M$, $I$ is bounded above. Thus,
$$\xi = \sup (I)$$
exists. We claim that $\Gamma (\xi )$ is boundary point of $M$. Observe that by continuity of $\Gamma$ it follows that $\xi \ne \max (I)$, that is, $\xi \notin I$.
Indeed, assume $\xi \in I$. Then $\Gamma (\xi )$ is interior point of $M$ and there is some $V^{\varepsilon} ( \Gamma (\xi ))$ symmetrical neighborhood of $\Gamma (\xi )$ contained in $M$. Since $\Gamma$ is continuous, there is some $U^{\delta}(\xi )$ neighborhood of $\xi $ such that $\Gamma (U^{\delta}(\xi )) \subset V^{\varepsilon} ( \Gamma (\xi ))$, which, in particular, implies that images of points $\gamma \in (\xi , \xi + \delta )$ are also interior points of $M$, which is a contradiction. Thus, $\xi$ being $\sup (I)$, cannot belong to $I$.
It remains to show that $\Gamma (\xi )$ is not exterior point of $\xi$. Assume $\Gamma (\xi )$ is an exterior point. Then, by the same reasoning as above, we conclude that theimages of points $\gamma \in (\xi – \delta , \xi ) $ are exterior points, from which follows that points $\gamma \in (\xi – \delta , \xi )$ are upper bounds to $I$, contradicting assumption that $\xi = \sup (I)$.
Hence $\Gamma (\xi )$ is neither interior point of $M$, nor exterior point of $M$, which leaves only possibility of $\Gamma (\xi )$ being boundary point of $M$ as claimed.
Best Answer
If $M$ is a any subset with interior $I:=\operatorname{int} M$, its exterior is $E:=\operatorname{int} M^\complement$. As in any metric space
$$\Bbb R^n = I \cup E \cup \partial M$$
where the three sets are pairwise disjoint, $I,U$ are open and $\partial M$ is closed.
So if we have a path $\gamma: [0,1] \to \Bbb R^n$ where $x_0=\gamma(0) \in I$ and $x_1=\gamma(1) \in E$ we cannot have that $\gamma[[0,1]] \cap \partial M = \emptyset$ because that implies that $\{I \cap \gamma[[0,1]], U \cap \gamma[[0,1]]\}$ would be a disconnection of the connected space $\gamma[[0,1]]$ by disjoint open sets, contradiction. So $\gamma[[0,1]] \cap \partial M \neq \emptyset$ and so there is some $t_0 \in [0,1]$ such that $\gamma(t) \in \partial M$.
So it holds in any topological space, and any subset $M$ of it, nothing is specific about $\Bbb R^n$ here; no metric is used as well. I only need that $\gamma$ is continuous and $[0,1]$ is connected, essentially.