I have been given a proof that the closed topologist's sine curve $\{{(t,\text{sin}(1/t)):t\neq 0\}}\cup \{{(0,t):-1\leq t \leq t\}}$ is connected.
First, we assume it is not connected. Let $I=\{{(0,t):-1\leq t \leq t\}}$ and $G=\{{(t,\text{sin}(1/t)):t\neq 0\}}$. Then there are disjoint, nonempty, open $U,V$ s.t. $I\cup G=U\cup V$. Then $I=(I\cap U) \cup (I\cap V)$, but $I\cap U$ and $I\cap V$ are open and $I$ is closed, so we have a contradiction.
I do not know how to see that $I\cap U$ and $I \cap V$ are open. I thought it is possible that they are closed? I think this issue is analogous to the unanswered question I have here: (Equivalent definitions of connectedness).
Best Answer
An easier way is to use that the closure of any connected set is again connected. The topologist's sine curve is the closure of the graph $\{(t, \sin(1/t)) \mid t > 0 \}$, which is path-connected (hence connected).
By the above argument, both $$\{ (t, \sin(1/t)) \mid t>0 \} \cup \{(0,t) \mid -1 \leq t \leq 1\}$$ and $$\{ (t, \sin(1/t)) \mid t<0 \} \cup \{(0,t) \mid -1 \leq t \leq 1\}$$are connected. Your topologist's sine curve is the union of these two connected sets, which intersection $\{ (0,t) \mid -1 \leq t \leq 1 \}$ is again connected.