Show that $\mathbb R^2$ \ {$(x_0,y_0)$} is connected For some $(x_0,y_0) \in \mathbb R^2$, prove that  $\mathbb R^2\setminus \{(x_0,y_0)\}$ is connected, using the fact that  $A_x =\{ (x,y) : y \in\mathbb R\}$ and $B_y = \{(x,y) : x\in\mathbb R\}$ are connected subsets of $\mathbb R^2$ for all $x,y \in \mathbb R$ and also  Let $\{X_i\}_{i\in I}$ be a collection of connected subsets of $\mathbb R$ such that $X_i \cap X_j \neq \varnothing $ for $ i,j \in I $ then $\cup_{i\in I}X_i$ is connected.
The "using the fact that..." is just a hint given which confuses me more. Any idea about how to prove it? Please help.
Thanks in advance!
 A: Proposition 0 If $U, V$ are two connected open sets and $U \bigcap V \ne \emptyset $ then $U\bigcup V$ is also a connected open set.
Dem Argue by contradiction, assume that there 2 open sets  $M,N$ such that $ \overline{M} \bigcap N = \overline{N} \bigcap M = \emptyset$ and $ M \bigcup N = U \bigcup V$.
If $U \bigcap M \ne \emptyset$ and $V \bigcap N \ne \emptyset$, $U\cap M$ and $U \cap N$ are clearly two connected open components of $U$, which contradicts the connectedness of $U$.
$ \Rightarrow U =M$ or $ U=N$
By the same argument, we have a similar conclusion for $V$, then we get the contraction by using the fact that $ U \cap V \ne \emptyset$
Proposition 1 The following sets $A=\{(x,y) : x>0\},B=\{(x,y) : <0\},C=\{(x,y) : y>0\}, D=\{(x,y) : y<0\}$ are connected open sets.
Dem Let $P(x,y)=( e^x,y )$ , it is a continuous surjective function between $R^2$ and $A$, therefore $A$ is connected by the fact that $R^2$ is connected. The rest can be obtained by the same argument.
Back to our problem, we apply the proposition 1 for the sets in proposition 2
A: Why not try to show it's even path connected. Take any $2$ points,say $a$ and $b$ in $\mathbf{R}^2-(x_0,y_0)$ . Look at the straight line joining them in $\mathbf{R}^2$, if $(x_0,y_0)\notin$ the line segment, we are done. If it is, draw a line at $b$ perpendicular to the line segment joining $a$ and $b$. Take any point on the perpendicular line other than $b$. Say, it is $c$. Now the straight line joining $b$ and $c$ does not pass through the point $(x_0,y_0)$ according to the construction, same goes for the straight line joining $a$ and$c$. To get a path from $a$ to $b$, first go from $a$ to $c$, then go from $c$ to $b$ via straight lines. Both lines are inside $\mathbf{R}^2-(x_0,y_0)$. So ,it is path connected.   
A: Let $C_{xy} = A_x\cup B_y$. It is connected by the hint you were given (explain!). Now, $$\mathbb R^2\setminus\{(x_0,y_0)\} = \bigcup_{(x,y)\in\mathbb R^2\setminus C_{x_0y_0}} C_{xy}$$ and you are done (explain!).
