Prove that if $τ'$ is a topology on $X$ for which the collection of $N_x$ is an open neighborhood system then $τ=τ'$. Definition 5: Let $(X,T)$ be a topology space , suppose $\forall x\in X$ we have a collection  $N_x$ of open sets with the following properties
i) $N_x \neq \emptyset$
ii) if $x \in N$ for $N$ open then $N \in N_x$
iii) if $N_1,N_2 \in N_x$ and $x \in N_1 \bigcap N_2$ then there exists a $N_3 \in N_x$ such that $x \in N_3$ and $N_3 \subset  N_1 \bigcap N_2$
iv) if $N \in N_x$ and $y \in N$, there exist an $N' \in N_y$ such that $N' \subset N$
v) $U \subset X$ is open if and only if $\forall x \in U$, there exist an $N \in N_x$ such that $N \subset U$
Proposition 7: Suppose $X$ is any set and for each $x \in X$ we have a collection $N_x$ of subsets of $X$ which satisfies i) through iv) of definition 5. Then if we define open subsets of $X$ by means of v) the set $T$ of open subsets is a topology on $X$ for which the collections $N_x$ are the neighborhood systems. 
Question
Suppose that $X$ is a set and that for each $x∈X$, we have a collection  $N_x$ of subsets of $X$ satisfying i) through iv) of definition 5.
Assume that the $N_x$ are used specify topology on $X$ in accordance with proposition $7$. Prove that if $τ'$ is a topology on $X$ for which the collection of $N_x$ is an open neighborhood system then $τ=τ'$.
this is what I got
Assume that the $N_x$ are used to specify a topology on $X$ in accordance with proposition $7$. Suppose that $τ'$ is a topology on $X$ for which the collection of $N_x$ is an open neighborhood system
From property iv) I got $τ' \subset τ$. I 'm struggle to prove the converse. 
 A: So you're left with proving that $\tau$ is the coarsest topology for which $\{ N_x \}_{x \in X}$ is a neighbourhood system.  I'll go slightly better: $\tau$ is the coarsest topology such that each set in each $N_x$ is open.
Hint: Start with an arbitrary topology $\tau^\prime$ in which each set in each $N_x$ is open.  Given any $\tau$-open set $U$, we know from the definition of the topology $\tau$ (this is essentially point (v)) that for each $x \in U$ there is a $V_x \in N_x$ such that $V_x \subseteq U$.  Now use properties of $\tau^\prime$ to show that $U$ is $\tau^\prime$-open.
A: There seems to be some error in the statement. Namely ii) in the definition. It seems to say that any subset containing the point $x$ should be in $N_x$, which is contradictory to the fact that $N_x$ is only allowed to contain open subsets, unless we are looking at the discrete topology. On the other hand merely adding that this subset should be open, somehow makes the proposition impossible since there would be a statement about open subsets while this concept is not yet introduced. 
It seem to me somehow more logical to read: if $N\in N_x$ then $x\in N$. So that the sets $N_x$ only contain "neighborhoods" of $x$. In fact this is needed for the proposition to hold, otherwise I might include some subset not containing $x$ in $N_x$, then the conditions
i) through iv) would still be satisfied, but $N_x$ could never be the neighborhood system for $x$. 
I find it hard to interpret the question correctly, as also Arthur mentions in the comments. The way I would rephrase it the proposition answers the question right away 
since $N_x$ is a neighborhood system and therefore any open in $\tau$ is in some $N_x$. (although -see the comment- in the current form the proposition does not hold).
