For a topological space $T=(X, \tau)$, then $A \subset X$ is $ \boxed{\quad\vphantom{A}\quad}$ if there is $\{\mathscr{T}_\alpha \} \subset \tau $ such that
for every $a \in A$ there is $\mathscr{T}_a \in \{\mathscr{T}_\alpha \}$ such that $a \in \mathscr{T}_a$, and
for every $a, b \in A$ with $a \neq b$, then $\mathscr{T}_a \bigcap \mathscr{T}_b = \emptyset$.
For example, under the usual topology with $X=[0,1]$, then
$ \{ \frac{1}{n+1} \mid n \in \mathbb{N} \}$ is $\boxed{\quad\vphantom{A}\quad}$, but
$\mathbb{Q} \cap (0,1)$ is not $\boxed{\quad\vphantom{A}\quad}$.
It implies $A$ is (relatively) discrete, i.e. discrete in the subspace topology. It's clear that $\mathcal{T}_a \cap A = \{a\}$. (BTW, "stylistically" it's better to name the sets $O_a$, say, as $\mathcal{T}_a$ will suggest a family of sets rather than just one set.)
If we have totally (i.e. also outside of $A$) disjoint $O_a$ as you demand, you could say that the $O_a$ are "simultaneously separated" (there is no standard term): it's the conclusion when a space is so-called collectionwise Hausdorff: a space is called that when every relatively discrete set $A$ has such a separating family of pairwise disjoint open sets.
E.g. all metric spaces have this property: they are paracompact (which implies collectionwise normal which implies collectionwise Hausdorff). So in a metric space all discrete sets are "simultaneously separated", and there is no real difference. In general a Hausdorff space need not be collectionwise Hausdorff, and there could be a discrete subspace without such a simultaneous separation. E.g. Bing's "example H" (discussed here) is a classic example of this.
Maybe calling $A$ "strongly discrete" would also be an option. I'm not sure if it's already taken or not. We could then say that $X$ is Collectionwise Hausdorff iff every discrete subspace is strongly discrete (or simultaneously separated). Just a thought.