Connected open subsets in $\mathbb{R}^2$ are path connected.
Solution 1:
The above proof is almost correct. The set $C$ should be the set of points that cannot be joined with $a$ by a continuous path within $U$. $C$ is open since every $y\in B_r(x)$ cannot be joined with $a$ for $r>0$ such that $B_r(x)\subset U$. Otherwise, we could join $a$ with $y$ and $y$ with $x$ by linear path so $x$ could be joined with $a$. That is a contradiction.