Is it a connected set in $\mathbb{R}^2$? I have this set $B=A_1\cup A_2\cup A_3\cup \{(0,0)\}$ where 
$A_1=\{(x,y)\in \mathbb{R}^2, y>x^2\}\\A_2=\{(x,y)\in \mathbb{R}^2, 0<y<x^2, x<0\}\\A_3=\{(x,y)\in \mathbb{R}^2, 0<y<x^2,x>0\}$
If i suppose that $B$ is not connected then there exist two open, disjoint sets $U,V$ in $B$ such that $B=U\cup V$ and i suppose that $(0,0)\in U$
If i draw $B$ i see that any disc with center (0,0) has an nonempty intersection with $A_i, i=1,2,3.$ 
But i can't find contradiction with the fact that $U$ is open or with $U\cap V=\emptyset$
Thank you.
 A: Yes, the union is connected and you have the correct idea. Since $(0,0) \in U$ and the $A_i$ are connected, and $A_i \cap U$ is nonempty, it must follow that $A_i \subset U$ for $1 \leq i \leq 3$. Try to prove this. But then $V$ must be empty.
This is a special case of a more general statement: if $X$ is a topological space and $\{C_i\}_{i \in I}$ is a family of connected subspaces such that $\bigcap_i C_i$ is nonempty, then $\bigcup_iC_i$ is connected (try to prove this). You're basically applying this  to the sets $C_i=A_i\cup\{(0,0)\}$ for $1 \leq i \leq 3$, to show that the union is connected.
You can also prove directly that the desired set is path-connected.
A: Your goal is to show that $V = \emptyset$.
For all $i=1,2,3$ you have
$$A_i = (A_i \cap U)\cup(A_i \cap V)$$
with $A_i \cap U \neq \emptyset$ and $A_i \cap U, A_i \cap V$ open. Since $A_i$ is connected, you have $A_i \cap V = \emptyset$.
Finally, $V = (A_1 \cap V) \cup (A_2 \cap V) \cup (A_3 \cap V) \cup (\{ 0\} \cap V) = \emptyset$.
A: For all elements $(x,y)\in B$ there is path that connects $(x,y)$ and $(0,0)$. 
This implies that $B$ is path-connected, and consequently is connected.

In the context of your question: let it be that $B=U\cup V$ where $U,V$ are disjoint, open and non-empty. 
Let $u\in U$ and let $v\in V$ and let $p:[0,1]\rightarrow B$ be a path with $p(0)=u$ and $p(1)=v$. 
Then $[0,1]=p^{-1}(U)\cup p^{-1}(V)$ where $p^{-1}(U)$ and $p^{-1}(V)$ are open, disjoint and non-empty. 
This however contradicts that $[0,1]$ is connected.
A: It is path-connected, hence connected. 
Sketch:
Indeed, for any point $M_i\in A_i\enspace (i\in \{2,3\}$, consider the arc $\gamma_M$ of the parabola that passes through and is tangent to the $x$-axis at $O$, and for a point $M$ in $A_1$, the line segment $[MO]$. Any points $M, N\in B$ can always be connected by the union of two of these paths.
