Use Neighbourhood system to recover a topological space, the "new" neighbourhood system is just the "old" one

Question

The following is about the axioms of a neighbourhood system of a topology.

How to prove the last sentence that the "new" neighbourhood system of $x$ is just the family of sets $\mathscr{U}_x$ we begin with.

My attempt:

1. To prove that the neighborhood system at each $x \in X$ is precisely $\mathscr{U}_x$. We prove these two families of sets contain the same sets.
2. $\Leftarrow$: If there is any open set $G$ containing $x$, by N-e) and N-d), $G$ itself qualifies as a member in $\mathscr{U}_x$ containing $x$. Then by N-d) again, every set including $G$ qualifies as a member in $\mathscr{U}_x$. So we have proved that every neighbourhood of x in the generated topology is a member of $\mathscr{U}_x$.
3. $\Rightarrow$. If now we have a member $U \in \mathscr{U}_x$(I emphasize here this is a just family of sets satisfying several axioms, are not guaranteed to be a neighbourhood system). I want to prove $U$ is indeed a neighbourhood of $x$, i.e. $U$ includes an open set $G$ containing $x$. How do I proceed further? In my guess, $U$ might just have an empty interior!

Here is a similar question, but the answerer didn't prove it.

Answer 1

First, note that, in (N-c), we can always assume that $V\subset U$ (if not, replace $V$ by $U\cap V$).

Start with a neighbourhood system $\langle \mathscr U_x:x\in X\rangle$ and let $\tau$ be the induced topology, so that $$ W\in\tau\iff \forall x\in W\exists U\in\mathscr U_x(U\subseteq W) $$

Let $\langle \mathscr N_x:x\in X\rangle$ be the neighbourhood system associated to $\tau$. More explicitly, $$ \mathscr N_x:=\{N\subset X:\exists U\in\tau(x\in U\subseteq N)\} $$

We need to prove that $\mathscr U_x\subseteq \mathscr N_x$ for every $x\in X$.

Fix $U\in \mathscr U_x$. Let $V\in\mathscr U_x$ be as in condition (N-c), with $V\subseteq U$.

Given $y\in V$, $U\in\mathscr U_y$, so (N-c) yields some $W_y\in\mathscr U_y$ with $W_y\subseteq U$ and $W_y\in\mathscr U_z$ for every $z\in W_y$. Put $$ W:=\bigcup_{y\in V}W_y\subseteq U $$ Since $x\in V$, $x\in W_x\subseteq W$.

I claim that $W\in\tau$. To see this, let $z\in W$, say $z\in W_y$ where $y\in V$. We know that $W_y\in\mathscr U_z$ and $W_y\subseteq W$. By definition of $\tau$, $W\in\tau$.

We've shown that $x\in W\subseteq U$ and $W\in\tau$. Therefore, $U\in\mathscr N_x$.