TITLE: Show that if $U$ is an open connected subspace of $\mathbb{R}^2$, then $U$ is path connected
QUESTION [5 upvotes]: Show that if $U$ is an open connected subspace of $\mathbb{R}^2$, then $U$ is path connected. (Hint:Show that given $x_0 \in U$, the set of points can be joined to $x_0$ by a path in $U$ is both open and closed in $U$.)
This should not be too difficult, but I am stuck at some point.
Let $x_0 \in U$
Let $A = \{ x \in U | \text{there is a path connecting $x$ and $x_0$} \}$ $\subset U$. 
I want to show that $A$ is open, so let $x \in A$. By openness of $U$, we can find a basic open set $\prod_{i=1}^2 (a_i,b_i)$ such that $\prod_{i=1}^2 (a_i,b_i) \subset U$. But I am stuck in showing that $\prod_{i=1}^2 (a_i,b_i)$ lies entirely in $A$.

REPLY [2 votes]: 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.