The closure of a subspace 
Let $X$ be a topological space. If $A$ is a subspace of $X$ we denote its closure by $\overline A$. For each point $x \in X$ the family $N_x$ of neighbourhoods of $x$ is a filter on $X$, the $\textit{neighbourhood filter}$ of $x$.

This is from Bell and Slomson (1969) Models and Ultraproducts: an introduction, p.22, who prove $\textit{inter alia}$ that, where $F$ is a filter on $X$, (i) and (ii) are equivalent: 
(i) $x \in \bigcap\{\overline A: A \in F\}$ 
(ii) For all $A \in F$ and $U \in N_x, A \cap U \neq \emptyset$
By $\overline A$ do they mean the intersection of all the closed sets of the subspace topology $(A, \tau_A$) that contain $A$; or, alternatively, do they simply mean the set of closed sets of the subspace topology? That is, suppose we have a topological space $(X, \tau)$ with $X = \{a, b, c, d, e\}$ and topology $\tau = \{X, \emptyset, \{a\}, \{d\}, \{a, d\} \}$, and suppose we have a subspace topology $(A, \tau_A$) of the topological space $(X, \tau)$, with $A= \{a, c\}$ and $\tau_A = \{A, \emptyset,\{a\} \}$. The closed sets of $(A, \tau_A$) are then: 
$A - A = \emptyset, \hspace{0.8cm} A - \emptyset = A, \hspace{0.8cm} A - \{a\} = \{c\}$
The intersection of all these closed sets containing $A$ is then $A$. So the closure by $\overline A$ in this case would be $A$. Is that right?
Or, alternatively, by the "closure by $\overline A$" do Bell and Slomson just mean the set of closed sets of $(A, \tau_A$), with $A= \{a, c\}$; i.e, the set containing: 
$A - A = \emptyset, \hspace{0.8cm} A - \emptyset = A, \hspace{0.8cm} A - \{a\} = \{c\}$

When you write $\overline{A}$ in your last sentence, do you mean the closure in the subspace topology? This would be A, as you said and your argumentation is correct. Or the closure in the topology of X? In this case: $\overline{A} = \{a,b,c,e\}$.
