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?