$A\in\overline{\tau}\iff\forall a\in A-E,\exists B\in\tau,a\in B\subset A$ defines a topology 
Let $\{X,\tau\}$ be a topological space. $E$ is a non-empty subset of $X$.
Show that it can be defined a topology $\overline{\tau}$, over $X$ in the following way:
$A\in\overline{\tau}\iff\forall a\in A-E,\exists B\in\tau,a\in B\subset A$

I know that the following criteria must comply for $\overline{\tau}$ to be a topology.
$\emptyset, X\in\overline{\tau}\\\bigcap_\limits{i\subset I}A_i\in\overline{\tau}\\\bigcup_\limits{i\subset I}A_i\in\overline{\tau}  $
If $A_1...A_i\in\overline{\tau}$, then $A$ is open. $\bigcap_\limits{i\subset I}A_i-E$ must have a $a\in \bigcap_\limits{i\subset I}B_i\subset \bigcap_\limits{i\subset I}A_i$
Question:
1) How do I prove this?
2) Am I going in the right direction?
Thanks in advance!
 A: $\overline{\tau}$ is just the topology generated by $\tau \cup \{A: A \subseteq E\}$, so the smallest topology that contains $\tau$ and makes all subsets of $E$ open.
For a direct proof that this is a topology:


*

*$\emptyset \in \overline{\tau}$, as we quantify universally over an empty set, so the definition holds and $X \in \overline{\tau}$ as we can take $B= X$ for all $a \in X\setminus E$.

*If $O_1, \ldots ,O_n \in \overline{\tau}$, and take any $a \in \cap_{i=1}^n O_i \setminus E$. Then $a \in O_i \setminus E$, so the definition of $\overline{\tau}$ gives us $B_i \in \tau$ such that $a \in B_i \subseteq O_i$.
Then $B= \cap_{i=1}^n B_i \in \tau$ (as a topology is closed under finite intersections) and $a \in B \subseteq \cap_{i=1}^n O_i$ as required. So $\cap_{i=1}^n O_i \in \overline{\tau}$.

*Suppose $I$ is an index set and for each $i \in I$ we have $O_i \in \overline{\tau}$. Define $O = \cup_i O_i$. To see $O \in \overline{\tau}$, pick $a \in O \setminus E$, so that for some $i$, we have $a \in O_i$, and so $a \in O_i \setminus E$, and the definition of $O_i \in \overline{\tau}$ gives us $B \in \tau$ such that $a \in B \subseteq O_i \subseteq O$, so that $B$ also shows that $O \in \overline{\tau}$.
