Let $O\subset\Bbb{R}^n$ be an open set. Prove that $O$ is connected if and only if every two points $x,y\in O, x\ne y,$ can be connected by a path in $O$ made up of a finite number of segments (a piecewise linear path).
I feel that one direction is straight forward: assume that there is a piecewise linear path from any $x,y\in O$, say $\varphi(t)$ such that $\varphi(a)=x, \varphi(b)=y, \varphi\in C([a,b],\Bbb{R}^n)$ then $O$ is path connected, hence connected.
I feel the other direction is a bit more complicated and I'm not sure how to approach it in order to be rigorous.
Hint: Let $x_0$ be an element of $O$. Consider the set of all elements in $O$ that have a piece wise linear path to $x_0$ (which is clearly not empty since you can take a ball around $x_0$). Show this set is both open and closed, thus must be all of $O$.
To see that the set must be all of $O$, let the set be $X$. If $X$ is open and closed, then $O=X\cup (O\setminus X)$ is a separation unless $X=\varnothing$ or $O\setminus X=\varnothing$. As stated above, $X\neq\varnothing$, hence $O\setminus X=\varnothing$, or equivalently, $X=O$ which is what we wanted to show.