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: 
$A - A = \emptyset, \hspace{0.8cm} A - \emptyset = A, \hspace{0.8cm} A - \{a\} = \{c\}$

$\overline A$ means the closure of the subset A within the space X.
"A is a subspace of X" should read "A is a subset of X".  
When viewing A as a space the expression "A is a closed subspace" means the subset A is closed within the larger space.
