Closed ball minus a point is connected Could someone help me to prove that if $B\subset\mathbb{R}^n$ ($n>1$) is a closed ball, then $B\setminus \{p\}$ is a connected set for all $p\in B$? 

You can solve this problem by reducing to dimension $2$. Firstly, any closed ball in $R^n$ is homeomorphic to the unit closed ball in $\mathbb R^n$, so it suffices to prove that the unit closed ball has the property you mention. Now, in the $\mathbb R^2$, you can draw the argument to see that the closed unit ball minus any point is path connected. 
Now, for the general case, given the closed unit ball $B$ in $\mathbb R^n$, with $n\ge 2$, a point $p$ that is removed from $B$, and two points $x,y\in B-\{p\}$, let $V$ be any plane that contains the origin as well as $x$ and $y$ (this is where $n\ge 2$ is used). The intersection $V\cap B$ is homeomorphic to the unit ball in $\mathbb R^2$. Thus, the intersection $V\cap (B-\{p\})$ is homeomorphic either to the unit ball in $\mathbb R^2$ or to the unit ball with a point removed. In either case, $x$ and $y$ can be connected by a path in the intersection (by the two-dimensional argument) and thus also in $B-\{p\}$. 
This argument shows that the property is essentially two-dimensional. 
