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.

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.
