Proving the connectedness of this subset 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?
 A: Your Method 2 works, so the answer is yes.
In Method 1 you falsely conclude that $B(c, \epsilon) \subset A \cup B$, when in fact it only follows that $B(c, \epsilon) \subset X = A \cup B \cup \{c\}$. I doubt that there is an easy way to avoid constructing a path, but maybe someone else has an idea.
What you could do in the specific case of $X = \mathbb R^2$ (or, similarly, a ball) is the following. The set $\mathbb R^2 \setminus \{c\}$ deformation retracts onto $S^1$, so has the same homotopy type. In particular, $\mathbb R^2 \setminus \{c\}$ is path-connected since $S^1$ is.
Edit: In general, you might want to do something like this. Let $B \subset X$ be a ball around $c$. Then $B \setminus \{c\}$ is path-connected (there your argument works). Since $X$ was path-connected (you use here again that $X$ is open), so is $X \setminus \{c\}$: Any path going through $c$ can be replaced by using connectedness in $B$.
A: Your method (2) might work if you provide more details, to make sure it works without the extra (implicit) assumption that $X$ is convex. The additional detail required may make that approach a bit long. 
Here is how method (1) could be completed. 
Using that $X$ is open take a small positive $\varepsilon$ such that the ball 
$E=\{p:dist(c,p)<\varepsilon\}$ is contained in $X$. 
Note that the deleted ball (or disk, since we work in the plane) $D=\{p:0<dist(c,p)<\varepsilon\}$ is (obviously) path-connected. You could use your proof (method 2), using line segments, since $E$ is convex. Any two points in $D$ could be connected by either a line segment missing $c$, or by a path consisting of two such line segments. 
Now take any two disjoint open sets $A$ and $B$ such that $X\setminus\{c\}=A\cup B$. 
Assume without loss of generality that $A\cap D\neq\varnothing$. Since $D$ is connected, we must have $D\subseteq A$. Let $C=A\cup\{c\}$. Then $C$ is open, disjoint from $B$, and $X=C\cup B$. Since $X$ is connected, $B$ must be empty. Therefore one cannot represent $X\setminus\{c\}$ as the union of two nonempty disjoint open sets $A$ and $B$, showing that $X\setminus\{c\}$ is connected. 
A: Let $\ c\in X\subseteq\mathbb R^2,\ $ where $\ X\ $ is open in $\ \mathbb R^2.\ $ (The case of
$\ x\in\mathbb R^2\setminus X\ $ is trivially true).
Let $\ G\ H\ $ be disjoint open subsets of $\ R^2\ $
such that $\ G\cup H=X\setminus\{c\}.\ $ Consider an
open ball
$$ B\ := \mathscr B(c, r)\ \subseteq X\qquad\qquad
      (\mbox{where}\quad r>0) $$
Then each circle $\ \mathscr C(c, t),\ $ where
$\ 0<t<r,\ $ is contained in one of $\ G\ H,\ $ (and is disjoint with the other one). By Dedekind axiom of real numbers we see that all these circles are contained in the same set $\ G\ $ or $\ H;\ $
say, in $\ G.$
Then $\ G\cup\{c\},\ $ i.e.
$$ G\cup\{c\}\ =\ G\cup B\ \subseteq\ X $$
is open; furthermore
$$ (G\cup\{c\})\ \cup\ H\,\ =\,\ X $$
while
$$ (G\cup\{c\})\ \cap\ H\,\ =\,\ \emptyset $$
This means that $\ X=G\cup\{c\},\ $ i.e.
$\ X\setminus\{c\}=G\ $ cannot be decomposed into two non-empty open disjoint subsets $\ G\ H.\ $ This means that $\ X\setminus\{c\}\ $ is connected.
Great!
