I would appreciate help with my (sketch of a) proof for the following problem:
Question:
Let $E\subset \mathbb{R^n}$ be an open connected open set. Show that for any $x,y\in E$ there is a finite sequence $$x_0=x,x_1,...,x_n=y\in E$$ such that the line segment from $x_i$ to $x_{i+1}$ is contained in E.
What I did: It's quite rough yet, but what I did was saying that if one of these segments is no fully in E, then there would be a point in E arbitrarily close to something not in E, which would mean that E is not open, reaching thus a contradiction.
It doesn't seem quite right though.
Choose some $x_0\in E$ and let $F$ be the set of $y\in E$ that can be reached from $x_0$ on a path $P_y$ consisting of finitely many line segments that lie in $E.$
(i). For any $y\in F,$ there is an open ball $B(y,r)\subset E$ with radius $r>0,$ and for any $z\in B(y,r),$ the union of $P_y$ with the line-segment from $y$ to $z$ is a suitable $P_z.$ So $B(y,r),$ which is open in the space $E,$ is a subset of $F$.
Therefore $F$ is open in the space $E$.
(ii).For any $y$ in the closure of $F$ in the space $E,$ we have $y\in E,$ so consider the same $B(y,r)$ as in (i). $ B(y,r)$ is a nbhd of $y$ in the space $E,$ and $y\in Cl_E(F),$ so there must exist $w\in F\cap B(y,r).$ Now the union of $P_w$ with the segment from $w$ to $y$ lies in in $E,$ so $y\in F$.
Therefore $F$ is closed in the space $E.$
(iii). By (i) and (ii), $F$ is a non-empty open-and-closed subset of the connected space $E,$ so $F=E.$ Now for any $x,y\in E,$ consider the union of the path $P_x$ from $x$ to $x_0$ with the path $P_y$ from $x_0$ to $y.$