I want to proof the following statement :
Let X be an open connected subspace of $\mathbb{R^2}$. Show that X is also path connected. The standard way to prove this problem from what I at least saw was to notice that a connected topological space has only two clopen subsets, namely X and the empty set. So if we define A as $A:=\{y\in X : \exists f: [0,1] \rightarrow X , f(0)=x, f(1)=y, \mbox{ f is continuous }\}$, the set which consists of points path connected with x. We can show that A is clopen and then A must equal X. I want to proof if it in an alternative way and for that I mainly use Munkres Theorem 25.5 :
Theorem 25.5:
If X is a topological space, each path component of X lies in a component of X. If X is locally path connected, then the components and the path components are the same.
If X is connected, then it has only one component. If I can show it is locally path connected then using this theorem we find that X has only one path component; X itself which means it is path connected. So far this is what I got :
Let $x\in X$ with X an open connected subspace of $\mathbb{R^2}$: An open neighbourhood of X is also an open neighbourhood of $\mathbb{R^2}$(by lemma 16.2). X is locally path connected at x if for every neighbourhood U there exists an open neighbourhood V such that V is path connected. Given any open neighbourhood of x , U, there exists $r>0$ such that V=$B_r(x)$, the open ball centered around x with radius r, is an open neighbourhood and V is a subset of U. An open ball in $\mathbb{R^2}$ is a circle which is convex and thus is path connected. We can do this in a similar fashion for all $x\in X$. Can we conclude that X is locally path connected? If not where is the fallacy in my argument? I would appreciate a hint or any kind of help.