I don't see the reasoning clearly behind the number of connected components in the following example taken from An Introduction to Manifolds by Loring Tu (First Edition, page no. 48).
My points of confusion are as follows.
In the Solution above I don't see why $U-\lbrace p \rbrace$ has four connected components.
The Solution doesn't mention the number of connected components of $B - \lbrace 0 \rbrace$ for $n \geq 2$. How can I determine it so that I can use the fact that the number of connected components is a topological property; i.e., preserved under homeomorphism to complete the solution?
In the first line of the Solution, we have assumed that $p \in U \cong B(0, \epsilon)$. Is that an assumption that the homeomorphism maps $p \in U$ to $0 \in B(0, \epsilon)$?
The answer to your third question is yes; the first sentence of the solution says explicitly that $p$ has a nbhd $U$ homeomorphic to an open ball $B(0,\epsilon)$ with $p$ mapping to $0$. Clearly the mapping referred to must be the homeomorphism taking $U$ to $B(0,\epsilon)$.
As for the first question, take $p$ to be the point where the arms of the cross come together: when you remove that point, the deleted cross that remains is the disjoint union of the four arms, each of which is homeomorphic either to $(0,1]$, if they include their far endpoints, or to $(0,1)$, if they do not. Each is therefore a connected set, and each is relatively open in the deleted cross, so they its connected components. Any open nbhd of $p$ must intersect each of them, so it must have at least four connected components. (It actually could have more, depending on the precise open set $U$.)
I don’t understand your second question, because the solution tells you exactly how many components an ball in $\Bbb R^n$ has after its centre point is deleted: it has one connected component (‘is ... connected’) if $n\ge 2$ and has two connected components if $n=1$.