In Rudin's Real and Complex Analysis, he give a motivation for the definition of a topology using open sets in a metric space. He says, for a metric space $X$, and for a set $\tau$ of sets $E \subset X$ if
$(A)$ $\tau$ is the set of opens sets of $X$ (defined using neighborhoods and interior points)
then
$(B)$ $\tau$ is a topology. i.e.
$\,\,\,\,\,\,\,\,(i)$ $\,\,\,\,\emptyset \in \tau$, $X \in \tau$
$\,\,\,\,\,\,\,\,(ii)$ $\,\,\,\,\tau$ is closed under countable intersections
$\,\,\,\,\,\,\,\,(iii)$ $\,\,\,\,\tau$ is closed under countable and uncountable unions.
I understand the direction $(A)\implies(B)$. But is it necessarily true that if $\tau$ is a topology on a metric space $X$, then $\tau$ is the set of open sets. i.e. $(B)\implies(A)$. Or similarly, does $(B)\implies (A')$ where
$(A')$ $\tau$ is a set of open sets of $X$. (not the set of open sets).
If $(B)$ does not imply $(A)$ or $(A')$ then why do we use $(B)$ as the topological definition for open sets if it permits sets other than open in a metric space?