Assume $X$ is a connected open subset of $\mathbb{R^2}$ show that $X$\ $\{c\}$, $c\in X$, is connected. I have thought two ways of proving this but none of which looked convincing to me.
Method 1: Assume for a contradiction that $X$\ $\{c\}$ is disconnected then $X$\ $\{c\}$ $= A \cup B$ for some $A,B$ open sets and $A\cap B=\emptyset$. Hence $X=A\cup B\cup\{c\}$. Furthermore, since $X$ is open, there exists $\epsilon>0$ such that $B(c,\epsilon)$\ $\{c\}$$\subset A\cup B$. Then I am not too sure how to continue this argument. I thought about using the fact that $c\notin A\cup B$ and the fact that $A\cap B=\emptyset$.
Method 2 (sketch): This way involves showing $X$\ $\{c\}$ is path connected. Take $x,y\in X$\ $\{c\}$. If $x,y,c$ are not collinear then it is trivial. Suppose $x,y,c$ are collinear then since X is open, there exists $z$ such that $x,y,z$ are not collinear. Then it is easy to find a path from $x$ to $z$ and another path from $z$ to $y$, compose two together we have find a path for any two points in $X$\ $\{c\}$. Lastly since path connectedness implies connectedness and hence it is shown?