This is a question that has been asked before , but now I came up with another proof so I would like to know if I am doing anything wrong.
Let $A$ be an open connected supspace of $\mathbb{R}^2$. Consider point $x \in A$. Let $B$ be a subset of $A$ such that it contains all the points with a path to $x$. Since $A$ is supspace of $\mathbb{R}^2$ and $B$ a subset of $A$, it follows $B$ belongs to the subspace topology so it is also open. Any point has a path to itself. So $B$ is not empty. Then $B\neq \phi$. Moreover we know $B \cap (A-B) = \phi$ and $B \cup (A-B) = A$. By similar reasoning as before, $A-B$ is also in the subspace topology so it is open. Since $A$ is connected, then it cannot have a separation. This implies $A-B=\phi$ and so $A=B$. We said $B$ is the set of all points that have a path to an arbitrary point $x \in A$. It follows $A$ is path connected.
Is this a valid proof?
Best Answer
Look at the chain-characterisation of connectedness I gave here, which is close to your idea. Take a cover of $O$ by open balls that stay inside $O$. This can be done obviously as $O$ is open in $\Bbb R^2$. If $x,y \in O$ we can apply the charcterisation to this cover and have a finite sequence of open balls (with some centre and radius) $B_1, B_2, \ldots B_n$ such that $x \in B_1$, $y \in B_2$ and $B_i \cap B_{i+1} \neq \emptyset$ for $i=1,\ldots, n-1$.
Now we can form a polygonal path (all straight line segments, as all open balls are convex, these all lie within a ball) from $x$ to some intersection point of $B_1 $ and $B_2$, next to some intersection of $B_2$ and $B_3$, and so on till the some last intersection point to $y$. So then we have a nice polygonal path from $x$ to $y$ entirely inside $O$. So $O$ is path-connected.
In general we show in this way that a locally path-connected connected space is path-connected.