The axiom of choice is not required. Let $\mathscr{B}=\{(\leftarrow,a):a\in A\}\cup\{(a,\to):a\in A\}\cup\{(a,b):a,b\in A\}\cup\{A\}\;,$ and let $\mathscr{T}=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;.$
(In case $A$ contains only one point, you do need to include $A$ in $\mathscr{B}$.)
You’re worried about showing that $\mathscr{T}$ is closed under arbitrary unions, so suppose that $\mathscr{V}\subseteq\mathscr{T}$; we want to show that $\bigcup\mathscr{V}\in\mathscr{T}$. For each $V\in\mathscr{V}$ let $\mathscr{B}_V=\{B\in\mathscr{B}:B\subseteq V\}\;;$ AC is not required to define this subset of $\mathscr{B}$.
Claim: $V=\bigcup\mathscr{B}_V$.
Proof: Clearly $V\supseteq\bigcup\mathscr{B}_V$. On the other hand, $V\in\mathscr{T}$, so by definition there is some $\mathscr{U}\subseteq\mathscr{B}$ such that $V=\bigcup\mathscr{U}$. Clearly $B\subseteq V$ for each $B\in\mathscr{U}$, so $\mathscr{U}\subseteq\mathscr{B}_V$, and therefore $V=\bigcup\mathscr{U}\subseteq\bigcup\mathscr{B}_V$, and it follows that $V=\bigcup\mathscr{B}_V$. $\dashv$
AC is not used here: we chose just one subset of $\mathscr{B}$, and its existence was guaranteed by the hypothesis that $V\in\mathscr{T}$.
Now let $\mathscr{H}=\bigcup\{\mathscr{B}_V:V\in\mathscr{V}\}\subseteq\mathscr{B}$. Then
$\begin{align*}\bigcup\mathscr{V}&=\bigcup\left\{\bigcup\mathscr{B}_V:V\in\mathscr{V}\right\}\\ &=\bigcup\bigcup\left\{\mathscr{B}_V:V\in\mathscr{V}\right\}\\ &=\bigcup\mathscr{H}\in\mathscr{T}\;. \end{align*}$
Again, we’ve not used AC anywhere. AC is also not required in the proof that $\mathscr{T}$ is closed under finite intersections and contains $\varnothing$ and $A$.