$\textbf{Prove : An open connected subspace of }$ $\mathbb{R}^2$ $\textbf{is path Connected.}$
I know there are other proofs of this statement here.BUt i want a proof verification and how to complete my proof in this particular way.So it is not a repost!!!
$\textbf{PROOF}$:
Let $A$ an open subspace of $\mathbb{R}^2$. Now let $x_0 \in A$ . I know there exist a nbh of $x_0$ Call it $B(x_o,ε) \subset A$ since $A$ is open.Now take a limit point of that ball That exist inside $A$ call it $y_o$ and it doesnt belong in the ball. That $y_o$ exists since if didnt belong in $A \setminus Β(x_o,ε)$ it must belong in $B(x_0,ε)$ Then $B(x_0,ε)$ is clopen in $A$ which is connected so contradiction. Now Since i know its existence connect $x_0$ with $y_o$ with the $f(x)=(1-t)x_0+t(y_o)$ Connect those points. Is continuous .Now do the same thing for $y_o$. And do that for all $x \in A$. and hence $A= \bigcup B(x,ε) $ for the right $ε>0$ and i have connected each center of the balls.
$$\textbf{Problem}:$$ I can do my described proof for the first open nbrh of $x_0$ . And get the first 2 nbrh.But on the third what if my $y_o$ is a limit point of my second nbrh in $A$ but belongs to a previous nbrh? Mean i might not take the whole $A$ after all.How do i surprass this problem? $$\textbf{So in other words how do i make sure every nbrh is not a subset of the previous one!!!}$$