Open and connected in $R^n$ revised I am trying to understand the following: If we have an open and connected set in $R^n$ then it can be connected with line segments parallel to the axes.
I managed to prove this:
 If a set $U$ is open and connected in $\mathbb{R}^n$ then we can prove it is polygonally connected(there is a path formed from line segments completely contained in $U$).
My question now is how would I modify the path such that the line segments remain in $U$ and they are now parallel to the axes?
I would very much appreciate some help.
Thank you!

First note that for any cube $C=[-r,r]^n\subseteq\mathbb R^n$ any point $c=(c_0, c_1,\ldots c_{n-1})\in C$ is polygonally connected to the center of $C$ along the axes.
$$(0,0,0,\ldots,0)\to(c_0,0,0, \ldots, 0)\to(c_0,c_1,0,\ldots0)\to\ldots\to(c_0,c_1,\ldots c_{n-1})$$
Let $G$ be any nonempty open connected set in $\mathbb R^n$ and let $a\in G$.
Now set $A=\{g\in G\mid\text{$g$ is polygonally connected to $a$ along the axes}\}$.
Note that for any any $b\in A$ there's a cube $C=b+[-r,r]^n\subseteq G$ because G is open, so $C\subseteq A$. This means $A$ is open.
And for any $b\in\overline A$ there is a cube $C=b+[-r,r]^n\subseteq G$, so there's also a point $c\in A\cap C$, so $b\in A$. This means $A$ is closed.
Therefore $A=G$, because it's a nonempty clopen subset of a connected set $G$.
