$\mathbb{R^2} \backslash (0,0)$ is connected Here is my proof of this -
Suppose $X = \mathbb{R^2} \backslash (0,0)$ is not connected.
$\implies$ There exists non-empty disjoint open subsets $U, V \subseteq \mathbb{R^2} \backslash (0,0)$ s.t. $U \cup V = X$.
$\implies \overline {U \cup V} = \overline X = X$
But $U, V$ open $\implies U \cup V \neq \overline {U \cup V}$
I.e. $U \cup V \neq X$. This is a contradiction, hence $X = \mathbb{R^2} \backslash (0,0)$ is connected.
Is this proof valid?
 A: Hint: Show that the space is ven more: path-connected. Of the circle having two given points as diameter endpoints at most one point can be missing, hence at least one arc is left intact.
A: Hint: use the fact that $X$ is connected iff
$
f:X\to \{0,1\} 
$ is continuous $\implies f$ is constant.
details:
Find a continuous path $\gamma:[0,1]\to X$ from $x$ to $y$:
$$
\gamma(0) = x, \gamma(1) = y
$$ (look at Hagen von Eitzen's post for an explicit construction).
$f\circ\gamma:[0,1]\to\{0,1\}$ is continuous, hence is constant: $f(x)=
f\circ\gamma(0) = f\circ\gamma(1)=f(y)$.

The equivalence with your definition:


*

*if $X=A\cup B$ with $A,B$ non empty open sets, then 
define $f(x) = 1_A(x)$. Then $f^{-1}(1), f^{-1}(0)$ are open, hence $f$ is continuous, and $f$ is not constant.

*If $f:X\to \{0,1\}$  is continuous: the choice $A=f^{-1}(1), B=f^{-1}(0)$ proves that $X$ is not connected.



NB: you can prove that for an open space there is equivalence between 
being connected and path connected.
NB2: the result remains true for $X-\{x_1,\dots x_n\}$ instead of $X-\{x\}$.
