How can I prove that any ball in $\mathbb{R}^n$ is connected? As the title follows, how can I prove that any ball in $\mathbb{R}^n$ is connected? or can you give me a hint?
I have some ideas but I'm not sure about them. I thank any help you can give me!
Regards 
 A: Suppose $B$ is not connected.  Then $B=C\cup D$ where $C$ and $D$ are both open in $B$ and $C\cap D=\emptyset$.  Let $p\in C$ and $q\in D$.  Define $f:[0,1]\rightarrow B$ by $f(t)=tp+(1-t)q$.  Then $f$'s components are polynomials in $t$ so is continuous.  But then $f^{-1}(C)$ and $f^{-1}(D)$ are two disjoint non-empty open sets of $[0,1]$ with $[0,1]=f^{-1}(C)\cup f^{-1}(D)$.  Thus if $B$ is not connected then $[0,1]$ is not connected.  But $[0,1]$ is connected (can we use this?).  QED
A: It is a consequence of the comb lemma:

If $C=A \cup\bigcup_{i\in I} B_i $, $A$ is non-empty and connected and
  for every $i\in I$ both $B_i$ and $A\cup B_i$ are non-empty and
  connected, then $C$ is connected.

Given that, you just have to consider a section $A$ of a ball through its centre and the set of perpendicular sections $B_i$s.
Once you know that the $1$-dimensional ball is connected, you may just use induction on $n$.
A: Product of connected sets is connected. Hence, the unit open cube is connected. The unit open cube is easily seen to be homeomorphic to the unit open ball. (For instance, both are homeomorphic to $\mathbb{R}^n$, since they both are open balls in some norm).  Now, every ball is homeomorphic to the unit ball.
If you are allowed to use path-connectedness, note that any ball is convex.
A: This follows from the fact that $[0,1]$ is connected (and path connected) and the fact that any ball is convex.
Since $\|tx+(1-t)y\| \le t\|x\|+(1-t)\|y\|$ for $t \in [0,1]$, we see that
any norm is convex, and it follows that any ball is convex.
Given any two points $a,b$ in a convex set $C$, we see that $\gamma(t) \in C$
for all $t \in [0,1]$ where $\gamma(t) = t a + (1-t) b$, and since $\gamma$ is
continuous, it follows
$C$ is path connected.
It follows from this that any convex set is connected. It is easy to see
by supposing that $U,V$ are two open disjoint sets such that $C \subset U \cup V$. If both are non empty, we can pick $a \in U, b \in V$ and let
$\gamma$ be the line joining them. Then $\gamma^{-1}(U), \gamma^{-1}(V)$ are
two non empty disjoint open sets containing $[0,1]$, which is a contradiction.
A: This is Aloizio Macedo's answer, with the words "cube" and "norm" removed.
Since the product of connected spaces is connected, $\mathbb R^n$ is connected.
Now, considering that
Open balls in euclidean space are homeomorphic to the whole space
we can conclude that any open ball is connected.
Since the closure of any connected set is connected, we also know that any closed ball is connected.
