Case 1
$B_1, B_2 \in \{D(z,\epsilon)\}$,
case (a) $B_1 \cap B_2=\phi$
case (b)$B_1 \cap B_2\neq \phi$
$x\in B_1 \cap B_2$.
Geometrically, I am able to see an element of $\{D(z,\epsilon)\}$,$B_3(say)$ such a way that $x\in B_3$ and $B_3\subset B_1 \cap B_2$. I took radius of $B_3$ as $dist(x, bound {(B_1\cap B_2)})/2$. Am I correct.
Case 2.
$B_1 \in \{D(z,\epsilon)\}$ and $B_2 \in \{E((x,0),\epsilon)\}$
Here also I choose radius of $B_3$ less than $dist(x, bound {(B_1\cap B_2)})$.
Case 3 $B_1,B_2 \in \{E((x,0),\epsilon)\}$
Here also I choose radius of $B_3$ less than $dist(x,\overline { bound {(B_1\cap B_2)})$.
Case 4
Do I left any case here? Is my idea of proof correct? Please help me.
Note in order to check a collection $\mathscr B$ of subsets of $A$ is a basis you have to check following two things
1) For each $a\in A$ there is at least one element $B$ containing $a$, which is trivially follow.
2) If $a$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $a$ such that $B_3\subseteq B_1\cap B_2$.
You have taken $a$ only from the set $\{(x,y)\in R^2 : y>0\}$ i.e. you have to consider also the case when $a\in \{(x,y)\in R^2 : y=0\}$.