Let $A$ be a subset of a topological space $X$ with a topology $T$. The following holds for $A$: For every $x\in A$ there exists an open set $U$ containing $x$ such that $U\subseteq A$.
Is it possible to construct a function $f:A\to B$ where $B=\{U\in T : U\subseteq A \}$.
Since neither $A$ nor $B$ are empty, it is always possible to construct a function. Fix some $U\in B$ and define $f(a)=U$ for all $a\in A$. But by the given condition, it seems that you want to somehow argue that $f(a)$ will be an open neighborhood of $a$, this is also easy, since the condition you give implies that $A$ is in fact open, so $f(a)=A$ is a suitable function.
If you want something a bit less trivial, then additional constraints are in order, for example, the function has to be injective. In that case, additional assumptions on the topology and the axiom of choice are necessary (e.g. if $T$ is the trivial topology, and $A$ contains more than one point, this is impossible).
Let me give an example where the axiom of choice actually plays a role. Suppose that there exists a subset of $\Bbb R$ which is infinite and Dedekind-finite, namely $D\subseteq\Bbb R$ is infinite but has no countably infinite subset. We may also assume that $D$ is dense, or at least unbounded.
Fix an enumeration of the rational numbers $\{q_n\mid n\in\Bbb N\}$ and define the following topology on the natural numbers: $U$ is open if and only if there is some $r\in D$ such that $U=\{n\mid q_n<r\}$, or if $U\in\{\varnothing,\Bbb N\}$. It is easy to verify two things:
Take $A=\Bbb N$ itself, if we want an injective choice function for open sets, then in particular it would mean there is a countably infinite set of open sets in the topology. But this is not the case, so this is impossible. Therefore any choice function has to have finitely many values.