A family $\displaystyle \mathcal{F}$ of sets generates a topology on $\displaystyle X$ just in case:
(1) $\displaystyle \mathcal{F}\subseteq 2^X$;
(2) $\displaystyle x\in X$ $\displaystyle \implies$ $\displaystyle \exists B\in\mathcal{F}$ such that $\displaystyle x\in B$;
(3) $\displaystyle x\in B_1\cap B_2$ for $\displaystyle B_1,B_2\in\mathcal{F}$ $\displaystyle \implies$ $\displaystyle \exists B_3\in\mathcal{F}$ such that $\displaystyle x\in B_3\subseteq B_1\cap B_2$.
Your collection satisfies these requirements. In fact, notice that the second term of the union is superfluous, and that $\displaystyle \mathcal{B}=\{B_d(x,r):x\in X,r>0\}$. Given that $\displaystyle X$ is finite, this means $\displaystyle \mathcal{B}=2^X\setminus\emptyset$, i.e. it generates the discrete topology.
I'm not sure what you mean by realizing a cover of a family of sets though.
Thank you for help!
Could you correct me if I am wrong in the following:
Let me illustrate examples geometrically.
Given X={a, b, c}
the topological base is
$\displaystyle \mathcal{B}$={{B_a},{B_b},{B_c},{B_a,B_b},{B_b,B_c}}
,
for
it is:
$\displaystyle \mathcal{B}$={{B_a},{B_b},{B_c}}
B_x - a ball for x