an open ball in $\mathbb{R^n}$ is connected 
Show that an open ball in $\mathbb{R^n}$ is a connected set.  

Attempt at a Proof: Let $r>0$ and $x_o\in\mathbb{R^n}$. Suppose $B_r(x_o)$ is not connected. Then, there exist $U,V$ open in $\mathbb{R^n}$ that disconnect $B_r(x_o)$. Without loss of generality, let $a\in B_r(x_o)$: $a\in U$. Since $U$ is open, for some $r_1>0$, $B_{r_1}(x_o)\subseteq U$. Since $(U\cap B_r(x_o))\cap (V\cap B_r(x_o))=\emptyset$, $a\not\in V$. Thus, $\forall b\in V, d(a,b)>0$. But then for some $b'\in V: b'\in B_r(x_o)$ and some $r>o$, $d(a,b')>r$. Contradiction since both $a$ and $b'$ were in the ball of radius $r$. 
Is this the general idea?
 A: An easy way to see this : a convex set is by construction path-connected, since $\forall x, y \in C$ with $C$ convex, $\lambda x + (1-\lambda)y \in C$ by convexity (so that you can choose the line between $x$ and $y$ as a path). Therefore since the unit ball is convex (show it if you wish to), and since path-connected sets are also connected, you're done. 
I'm not quite sure your proof is correct though. Why would $d(a,b') > r$? I don't see an explicit reason for this.
Hope that helps,
A: As mentioned, the simplest way to see this is via path connectedness.  However, I think induction works too:
We already know that open balls in $\mathbb{R}$ (i.e. open intervals) are connected.  Suppose that open balls in $\mathbb{R}^{n-1}$ are connected.  Consider an open ball $B(x,r)$ in $\mathbb{R}^n$, and view it as
$$B(x,r)=\{(x_1,\ldots,x_{n-1},y):|x_n-y|<r\}\cup\bigcup_{\varepsilon\in(x_n-r,x_n+r)}\{y\in B(x,r):y_n=\varepsilon\}.$$ Now $\{(x_1,\ldots,x_{n-1},y):|x_n-y|<r\}$ is an interval (and is then connected) and each $\{y\in B(x,r):y_n=\varepsilon\}$ is homeomorphic to an open ball in $\mathbb{R}^{n-1}$ (so that each such set is then connected).  Furthermore $\{(x_1,\ldots,x_{n-1},y):|x_n-y|<r\}\cap\{y\in B(x,r):y_n=\varepsilon\}\neq\emptyset$ for each $\varepsilon\in(x_n-r,x_n+r).$  There is a theorem (likely in your textbook) that now allows you to conclude that $B(x,r)$ is connected.
In particular the theorem in question states that $X\cup\bigcup_\alpha Y_\alpha$ is connected if $X$ is connected and each $Y_\alpha$ is connected, and furthermore $X\cap Y_\alpha\neq\emptyset$ for each $\alpha.$  This is a straightforward corollary of the more basic theorem that $\bigcup_\alpha Y_\alpha$ is connected if each $Y_\alpha$ is connected and $Y_\alpha\cap Y_\beta\neq\emptyset$ for each $\alpha,\beta$.
A: The last line, "Contradiction since both a and b′ were in the ball of radius r" is NOT a contradiction. It is only a contradiction if one of the points is the center of the ball, otherwise the farthest two points can be is 2r.
A: Hint: An open ball in $\mathbb R^n$ is convex, as can be shown with the triangle inequality. And convex set is path-connected, thus connected.
A: $\mathbb{R}=(-\infty,\infty)$, hence it is connected. Since the finite product of connected space is connected, the result follows.
