A path connected between the interior and exterior of a set. Let $(X,\tau)$ be a topological space and $A\subset X$. Prove that any path joint the interior of $A$ with the exterior of $A$ it's found in the $Fr(A)$. 
Proof: 
Let $x\in \mathrm{int}(A), y\in \mathrm{ext}(A)$ and $f:[0,1]\rightarrow X$ a continuous function such that
$f(0)=x$ and $f(1)=y$. 
I must prove that $f(t)\in Fr(A)=\overline{A}\cap (\mathrm{int}(A))^C$ for some $t\in (0,1)$. (Is this correct?)
I have that $f(0)\in \mathrm{int}(A)$, i.e., exist $V_1\in\tau$ a neighborhood of $f(0)$ such that 
$V_1\subset A$. 
Similarly, using the fact that $f(1)\in \mathrm{ext}(A)$, there exists $V_2\in\tau$ a neighborhood of $f(1)$ such that 
$V_2\subset A^C$. 
Any idea to proof that $f(t)\in Fr(A)=\overline{A}\cap (\mathrm{int}(A))^C$ for some $t\in (0,1)$?
 A: Note that the interior, boundary and exterior form a partition of $X$. This is the key fact.
We know that $f(0)$ is in the interior and $f(1)$ is in the exterior.
Let $T= \sup \{ t \in [0,1] | f(\tau) \in A^\circ \text{ for all } t \in [0,t] \}$.
It is straightforward to show that $f(T) \notin A^\circ$.
Show that $T \in (0,1)$ and $f(T) \in \partial A$ (otherwise...).
A: Let $f: [0,1] \to X$ be a (continuous) path where $f(0) \in \operatorname{int}(A)$ and $f(1) \in \operatorname{ext}(A)$.
Standard fact: $X$ is the pairwise disjoint union of the two open sets $\operatorname{int}(A)$, $\operatorname{ext}(A)$ and the closed set $\operatorname{Fr}(A)$ (the boundary of $A$). If $f[[0,1]] \cap \operatorname{Fr}(A) =\emptyset$ we thus would have that
$$[0,1] = f^{-1}[\operatorname{int}(A)] \cup f^{-1}[\operatorname{ext}(A)]$$
and these sets are non-empty, open and disjoint so this would contradict the connectedness of $[0,1]$. Hence $f[[0,1]] \cap \operatorname{Fr}(A) \neq \emptyset$, as asked for.
