I am working on Exercise 1.1 and I think the way to do this would be to show that open sets are homeomorphic to $\mathbb{R}^n$ or open balls in $\mathbb{R}^n$. Is this even true? I'm not sure how to go about proving it.
BTW: The exercise is from Lee's Introduction to Smooth Manifolds.
Open balls in $\Bbb R^n$ are homeomorphic to $\Bbb R^n$, but it’s not true in general that (non-empty) open sets in $\Bbb R^n$ are homeomorphic to $\Bbb R^n$: $\Bbb R^n$ and its open balls are connected, but there are lots of open sets in $\Bbb R^n$ that are not connected. However, if $U$ is an open nbhd of $x$ in $\Bbb R^n$, then there is an open ball $B$ such that $x\in B\subseteq U$, so if every point $M$ has a nbhd homeomorphic to some open $U\subseteq\Bbb R^n$, then it automatically has one homeomorphic to an open ball in $\Bbb R^n$. The other direction is trivial, since every open ball in $\Bbb R^n$ is an open set in $\Bbb R^n$.
Finally, to prove that an open ball in $\Bbb R^n$ is homeomorphic to $\Bbb R^n$ itself, it suffices to prove it for the open unit ball centred at the origin. Consider the map from the open unit ball to $\Bbb R^n$ that sends $x$ to $\left(\tan\frac{\pi|x|}2\right)x$.