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Argument strengthened to provide a more definitive answer, removed unnecessary sentence

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 $\tau_\nu=\nu\cup\{\varnothing,X\}$. More precisely:

  • Since $\nu$ is closed under finite intersections, we conclude that $\nu\cup\{X\}$ is even a basis of $\tau_\nu$. This implies that $\tau_\nu$ is the unique topology on $X$ for which $\nu\cup\{X\}$ is a basis and thus the coarsest topology on $X$ containing $\nu$;

  • Since $\nu$ is closed under taking supersets, we conclude that any open set $U\in\tau_\nu$, being an union of members of $\nu\cup\{X\}$, belongs itself to $\nu\cup\{X\}$. Therefore, $\nu\cup\{\varnothing,X\}$ satisfies all the axioms for a topology on $X$. Since $\tau_\nu$ is the coarsest topology on $X$ containing $\nu$, we must have $\tau_\nu=\nu\cup\{\varnothing,X\}$, as claimed.

The property $\bigcap_{U\in\nu}U=\varnothing$ dual to a bornology being a cover of $X$ amounts to $\tau_\nu$ being $T_0$.

Remark: if you were dealing with just Boolean algebras instead of (bounded) lattices, then we would have to require that $X\in\nu$. Moreover, proper filters must satisfy $\varnothing\not\in\nu$, so including $0=\varnothing$ and $1=X$ to $\tau_\nu$ by hand is necessary in the context of (bounded) lattices.

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.