Let $U$ be an open connected subset of $\mathbb R^2$, $x_0\in U$, and $$S=\{x\in U: \text{there exists a path from $x_0$ to $x$}\}.$$ We show that $S$ is clopen in $U$, and because $U$ is connected, is therefore equal to $U$. First let $x\in S$, then because $U$ is open, there exists a neighborhood (i.e. open ball) $V$ of $x$ with $V\subset U$. Since neighborhoods are path connected, there exists a path from $x$ to $y$, so by the pasting lemma there exists a path from $x_0$ to $y$. Hence $V\subset S$, so that $S$ is open in $U$.
Now let $x$ be a limit point of $S$. Then each neighborhood $V$ of $S$ intersects $S$ in a point $y\ne x$. Since $y\in S$, there exists a path from $x_0$ to $y$, and since $V$ is path connected, there exists a path from $y$ to $x$. So again by the pasting lemma, there exists a path from $x_0$ to $x$. It follows that $x\in S$, so that $S$ is closed in $U$.
I don't see any error in my proof, but I did not use the fact that $U$ is open in showing that $S$ is closed, so I am unsure of its validity.
As an aside, this same argument (provided it is correct) can be used for $\mathbb R^n$, right?