A region in $\mathbb C$ minus countably many points is path-connected Edit: Someone marked my question as a duplicate of this. However, I am asking about proving this result for any open and connected set in the plane and when removing countably many points and not necessarily finite. I know how to prove it when removing finitely many points. Please reopen the question.
I want to prove that if $\Omega$ is a region in the complex plane and if $A$ is a countable subset then $\Omega - A$ is connected. Given $z_1$ and $z_2$ in $\Omega$ we must prove that there is some path in $\Omega$ joining $z_1$ to $z_2$.
Now in case $\Omega$ is the whole complex plane then this can be done by taking a line passing through $z_1$ which does not intersect $A$ and another passing by $z_2$ but neither parallel to the first line nor intersecting A. These two exist since at each point there are uncountable many such lines since if all lines, for instance passing by $z_1$, intersect $A$ then this would contradict the countability of $A$. Let $z_3$ be the intersection of the two lines and thus join $z_1$ to $z_2$ by joining $z_1$ to $z_3$ and then $z_3$ to $z_2$.
I am not sure how to prove the more general case. If we join $z_1$ to $z_2$ by a path $\gamma$ in $\Omega$, then we need to modify this path so that it does not pass by A. Any suggestions how?
 A: You can do a similar argument as in the case of $\Bbb C$. The important trick is to reduce to the following case:

Lemma. Let $X$ be a countable subset of the unit ball $B_1(0)$. Then $B_1(0)-X$ is path connected.

Proof:
Let $x,y\in B_1(0)-X$. For $z\in B_1(0)-X$ let $[x,z]$ and $[y,z]$ denote the paths $\{t x+(1-t)z\mid t\in[0,1]\}$ and $\{ty+(1-t)z\mid t\in[0,1]\}$. These paths are contained entirely $B_1(0)$ and if $z_1\neq z_2$ with $x-z_1$ and $x-z_2$ not proportional to each other then $[x,z_1]\cap [x,z_2] = \{x\}$.
So, since $X$ is countable, there are uncountably many $z$ for which both $[x,z]\cap X$ or $[y,z]\cap X$ are empty. Pick one $z$ for which it is empty and then $[x,z]\cup [y,z]$ gives a path from $x$ to $y$ in $B_1(0)-X$.
The above obviously also implies:

For any $\epsilon >0$ and $x\in \Bbb C$ and countable $X$ the ball $B_\epsilon(x)-X$ is path connected.

Now to show the statement you are interested in. $\Omega$ is open and connected, so it is path connected. Let $x,y\in \Omega-X$ and let $\gamma:[0,1]\to \Omega$ be a path connecting $x$ and $y$. Since the image of this path is compact and $\Omega$ is open there are $x_1,...,x_n$ and $\epsilon_1,...,\epsilon_n$ so that
$$\gamma([0,1])\subset\bigcup_{k=1}^n B_{\epsilon_k}(x_k)\subseteq \Omega$$
Now each of the $B_{\epsilon_k}(x_k)-X$ are path connected and the intersection of two neighbouring balls is non-empty. Hence their union is path connected and there is a path connecting $x\in \bigcup_{k=1}^n B_{\epsilon_k}(x_k) -X$ to $y$ which lies in the same set.
