Show that if $U$ is an open connected subspace of $\mathbb{R}^2$, then $U$ is path connected
Solution 1:
Let $x\in A$. Then take an open ball $B$ containing $x$ small enough such that $B\cap U = B$. We know that open balls are path connected (you can easily construct a path), so $B\subseteq A$, and $A$ is open. To show that $A$ is closed, consider $x\in U\setminus A$ and show that there exists an open ball $B$ containing $x$ such that $B\cap A = \emptyset$ (hint: what would happen if $p\in B\cap A$?)
Solution 2:
To show $A$ is closed:
$x\sim y$ iff $x$ and $y$ are path connected
is an equivalence relation on $U$.
$U-A$, as a union of open subsets, is therefore open.