A path connected proof in $\mathbb{R}^2$ I'm trying to prove that if $U \in \mathbb{R}^2$ is open and path connected, then for a point $ p \in U$ we have $U \smallsetminus \{p\}$ still path connected.
Start by taking $x,y \in U$. As path connected there exists continuous $\gamma : [0,1] \to U$ such that $\gamma (0) = x$ and $\gamma(1) = y$.
Take $p \in U$. 
If $p$ is not on the the path $\gamma$. Then $x$ and $y$ are still path connected.
If $p$ is lying on the path $\gamma$ then as $U$ is open, there exists a $\delta >0$ such that the ball $B(p,\delta) \subset U$. The path $\gamma$ crosses the boundary of this ball. Now one can create a new path which traverses round the edge of the ball. Hence $x$ and $y$ are still path connected. 
Is this proof valid? Can anyone think of a simpler way of doing it? 
 A: Your idea is good, but the proof is not rigourous enough.


*

*You should choose $\delta>0$ such that $\delta < \min\{\|x-p\|, \|y-p\|\}$ and $B(p,c\delta)\subset U$ for some $c>1$. The first condition is needed because you want to keep $x$ and $y$ outside the open disc. The second condition is needed, otherwise the boundary of $B(p,\delta)$ may not lie inside $U$.

*As Chris Eagle mentioned in his comment, the original path $\gamma$ may cross the closed disc $\bar{B}(p,\delta)$ multiple times. So you should consider the first entry point $t_1 = \inf\{t: \|\gamma(t)-p\|\le\delta\}$ and the last exit point $t_2 = \sup\{t: \|\gamma(t)-p\|\le\delta\}$. You should show that these infinum and supremum are attainable minimum and maximum. Then you can remove the part of $\gamma$ on $[t_1,t_2]$ and replace it by a circular arc on the boundary of $B(p,\delta)$.

A: This is a comment: what if one of or both $x, y$ are inside the open nbd you have picked?
added How about this: Pick a point $q$ different from $p$ and consider the set of points in $U-\lbrace p \rbrace$ that can be path connected to $q$. This set and its complement are both open. If the complement is non-empty then it will provide a separation for $U$.
