As you have defined it, $B$ does not form a basis for a topology on $\mathbb{R}$. For $a<b<c$, we have $[a,b],[b,c]\in B$, but $\{b\}=[a,b]\cap[b,c]$, and no basis element both contains $b$ and belongs to $\{b\}$
As they told you in the comments, open discs form a basis for the euclidean topology in $\Bbb R^2$ by definition, since open sets in that topology are defined to be the union of open discs.
A basis $\mathcal B$ in a topological space $(X,\tau)$ is a collection $\mathcal B$ of open sets for the topology of $\tau$, $\mathcal B\subseteq\mathcal P(X)$, such that every element of the topology (every open set) can be expressed as a union of elements of the basis. Since this is the definition of basis, and the open sets of the euclidean topology of $\Bbb R^2$ are the union of open discs, it follows trivially that open discs form a basis.
There is an equivalent definition for basis, the local definition of basis: a basis is a collection $\mathcal B\subseteq\mathcal P(X)$ of open sets such that given any open set $U$ of the topology, and given any element $x\in U$, there is an element of the basis $B\in\mathcal B$ such that $x\in B\subseteq U$, this is, $B$ contains $x$, but it remains enclosed in $U$. As I said, the two definitions are equivalent. It would be a good exercise to prove this.
Now, it remains trivial that open discs form a basis in the euclidean topology of $\Bbb R^2$, since every open set $U$ was the union of open discs, so for any $x\in U$ there is an open disc from that union containing $x$ that's enclosed in $U$ (since all the union was $U$).
If closed discs formed a basis, they would be open sets, so their finite intersection woulb be open sets too. But you could take two closed discs in $\Bbb R^2$ touching the other one in exactly one point, for example the closed discs with radii 1 centered at $(0,0)$ and $(0,2)$; but then the intersection would be that point, so that point should be an open set, although there is no way you can express a point a union of open discs. Then we can't have the closed discs as a basis for the euclidean topology in $\Bbb R^2$.
Best Answer
Fix $a \in \mathbb{R}$. Let $\mathscr{B} = \{D((a,0), r) : r > 0\}$. Then $\mathscr{B}$ is a base, but it is not the base for the usual topology on $\mathbb{R}^2$.
To see that it's a base, we check that it covers $\mathbb{R}^2$ and that for any $B_1, B_2 \in \mathscr{B}$, there is a $B_3 \in \mathscr{B}$ with $B_3 \subset B_1 \cap B_2$.
For any $(x,y) \in \mathbb{R}^2$, let $r = d((x,y), (a,0))$. Then $D((a,0), r+1)$ is a basis element that contains $(x,y)$. So it covers $\mathbb{R}^2$.
Let $B_1 = D((a,0), r_1)$ and $B_2 = D((a,0), r_2)$ be two basis elements. Then $B_1 \cap B_2 = D((a,0), r')$ where $r' = \min(r_1, r_2)$. For any $(x,y) \in B_1 \cap B_2$, it will have distance less than $r'$ from $(a,0)$. Denote it $s$. Let $r_3 = \frac{r' + s}{2}$. Then $B_3 = D((a,0), r_3)$ will be a basis element containing $(x,y)$ and will be contained in the intersection.
This base does not generate the usual topology on $\mathbb{R}^2$. For example, consider the open disc of radius $1$ centered at $(0,2)$. You cannot write that open set as a union of basic open sets from your base.
To your edit: The answer is no, that this does not form a basis. Consider the discs centered at $(0,0)$ and $(1,0)$ with radii $\frac{2}{3}$. Then there will be a point in the intersection that has distance further than $\frac{1}{3}$ from the $x$-axis. Therefore, you cannot have any disc inside their intersection that contains that point.