It was not clear to me at first what your question has to do with bornologies, but now (**EDITED**) I see it. *Any* collection $\nu$ of subsets of $X$ is a subbasis of a *unique* topology $\tau_\nu$ on $X$ - to wit, the intersection of all topologies on $X$ containing $\nu$. The fact that $\nu$ is a filter will ensure that $\tau_\nu=\nu\cup\{\varnothing\}$. More precisely:

 - Since $\nu$ is closed under finite intersections, we conclude that $\nu$ is even a basis of $\tau_\nu$. This implies that $\tau_\nu$ is the unique topology on $X$ for which $\nu$ is a basis and thus the *coarsest* topology on $X$ containing $\nu$;
 
 - Since $\nu$ is closed under taking supersets (which, by the way, entails the fact that $1=X\in\nu$, implicitly used above), we conclude that any nonvoid open subset $U\in\tau_\nu$, being an union of members of $\nu$, belongs itself to $\nu$. Therefore, $\nu\cup\{\varnothing\}$ satisfies all the axioms for a topology on $X$. Since $\tau_\nu$ is the coarsest topology on $X$ containing $\nu$, we must have $\tau_\nu=\nu\cup\{\varnothing\}$, as claimed.

The property $\bigcap_{U\in\nu}U=\varnothing$ dual to a bornology being a cover of $X$ amounts to $\tau_\nu$ being $T_0$. On the other hand, $\nu=$ the collection of all nonvoid open subsets of $X$ being closed under taking supersets is considerably stronger than just being closed under unions, so this $\tau_\nu$ may be a *very* fine and disconnected topology.

**Remark:** proper filters must satisfy $\varnothing\not\in\nu$, so including $0=\varnothing$ to $\tau_\nu$ by hand is necessary.

If $X$ is a *vector* space (over $\mathbb{R}$ or $\mathbb{C}$) and $\beta$ is a *convex* bornology (i.e. closed under addition, scalar multiplication and absolutely convex hulls), there is a different way to define a topology on $X$ - namely, you adopt as your neighborhood filter of zero the collection of all $\beta$-*bornivorous disks* (i.e. absolutely convex subsets that absorb all members of $\beta$). This defines a locally convex topology on $X$ - the *finest* such one w.r.t. which the bounded subsets are precisely the members of $\beta$. However, it is clear that the vector space structure of $X$ plays an essential role in this case.

The duality between (convex) bornology and (locally convex) topology acquires a deeper meaning in the theory of locally convex vector spaces, since compatible topologies on dual spaces are defined in terms of bornologies of the original spaces, and vice-versa. The book "Bornologies and Functional Analysis" by Henri Hogbe-Nlend (North-Holland, 1977) discusses such topics in depth.