I am trying to solve the first exercise in John Lee's Introduction to Smooth Manifolds and I am confused by the terminology in the question.
He says (paraphrased): Consider the usual definition of a topological manifold $M$ (ie Hausdorff, second countable, and locally homeomorphic to $\mathbb{R}^{n}$). Show that equivalent definitions of manifolds are obtained if instead of requiring an open subset in $M$ to be homeomorphic to any open subset of $R^{n}$, we require it to be homeomorphic to an open ball in $\mathbb{R}^{n}$, or to $\mathbb{R}^{n}$ itself.
What kind of ball does he mean? Does he mean $n$-ball? Is this the same as an open set?
Here, an open ball probably refers to something like an open $n$-ball under the Euclidean norm on $\Bbb R^n$. No, not all open sets in $\Bbb R^n$ are balls, though all balls are open sets. However, f every $p \in M$ has a neighborhood that is homeomorphic in $\Bbb R^n$, then it follows that every point has a neighborhood that is homeomorphic to an open ball (the converse holds trivially).
In particular, suppose $p \in U \subset M$ with $U$ open, and $h: U \to V \subset\Bbb R^n$ is a homeomorphism.
By the definition of openness in $\Bbb R^n$, there exists an open ball $N$ around $h(p)$ such that $h(p) \in N \subset V$. From there, it follows that $p \in h^{-1}(N) \subset M$. That is, $p$ has neighborhood $h^{-1}(N)$, which is homeomorphic to an open ball in $\Bbb R^n$.