Definition: A subset $S$ of a topological space $X$ is said to be connected if there do not exist two open sets $U$ and $V$ such that $S\cap U\cap V =\emptyset$, $S\subset U \cup V$, and $U\cap S\neq \emptyset \neq V \cap S$. If such $U$ and $V$ do exist, then $S$ is said to be disconnected.
Problem: Show that the set $A=\{(x,y)\in \mathbb{R}^2: x^2+y^2<1 \}$ is connected using the previous definition.
Suppose $U,V$ are suitable sets and $a \in U\cap S$, $b \in V \cap S$, then let $p(t) = (1-t)a+tb$. Note that $p(t) \in A$ for all $t \in [0,1]$.
Let $t^* = \sup \{ t\in [0,1] | p(s) \in U \text{ for all } 0 \le s \le t\}$. Note that $t^* <1$ and $p(t^*) \notin U$ since $U$ is open, hence $p(t^*) \in V$. However, if $p(t^*) \in V$ then we must have $p(t) \in V$ for some $t < t^*$ which contradicts the definition of $t^*$. Hence no such $U,V$ exist.