Fund. system of neighborhoods/ Openness of set in topological space Let $X\neq\emptyset$ be a set and let $\mathfrak{B}_x$ be the fundamental system of neighborhoods of $x\in X$. The collection
$$\tau = \left\lbrace U\in\tau : (\forall x\in U)(\exists B_x\in\mathfrak{B}_x)(B_x\subset U)\right\rbrace $$
is a topology on $X$. Given the properties:
(1) $B_x\in\mathfrak{B}_x\Longrightarrow x\in B_x$
(2) $B^1_x,B^2_x\in\mathfrak{B}_x\Longrightarrow \exists B^3_x\in\mathfrak{B}_x : B^3_x\subset B^1_x\cap B^2_x$
(3) $B_x\in\mathfrak{B}_x\Longrightarrow\exists B^1_x\in\mathfrak{B}_x : B^1_x\subset B_x$ and  $\forall y\in B^1_x : \exists B_y\in\mathfrak{B}_y : B_y\subset B_x$  
Fix $x\in X$

Show that $V := \left\lbrace z\in B_x : (\exists B_z\in\mathfrak{B}_z)(B_z\subset B_x) \right\rbrace$ is $\tau$-open.

Attempt
Because $B_x\subset B_x$, we have $x\in V$ hence $V\neq\emptyset$. Let $z\in V$, to show $V\in\tau $, we must find $B_z\subset V$.  
I can't see any direct way to do this. For instance, there is no reason for any intersection of the sets $B_z$ to be wholly contained in $V$
Assume for a contradiction than $V\notin\tau$, then there exists $z\in V$ such that for any $B_z\in\mathfrak{B}_z$ $B_z\not\subset V$  
Again, seems fruitless
What I can say is that for any $n\in\mathbb{N}$ there exists some $B\subset \bigcap_{j=1}^n B^j_z$ per property (2). So, I could construct:
$$B_z\supset B^1_z\supset\ldots\supset B^m_z\supset B^{m+1}_z\supset\ldots  $$
And per our assumption all of these $B^j_z\not\subset V$.  
What I can't say, what happens as $m\to\infty$.  
Right now, I am hoping to arrive at $z\notin V$ which would be a contradiction.  
What can we do to make progress? Can we prove it directly?
 A: I'm not quite sure I understand the question, however I will attempt to answer it as best I can, namely, I will answer the following question, which is my interpretation of OP's question.

Let $(X,\tau)$ be a topological space. For every $x \in X$, let $\mathfrak{B}_x$ be a neighborhood basis of $x$. Let $x \in X$, and let $N$ be a neighborhood of $x$. Define
   $$
 V := \left\{z \in N\ |\!:\ \exists B \in \mathfrak{B}_z,\ B \subseteq N\right\}.
 $$
   Show that $V \in \tau$.

Proof
For every $z \in V$ denote by $B_z$ a member of $\mathfrak{B}_z$ such that $B_z \subseteq N$, and denote by $G_z$ an open set such that $z \in G_z \subseteq B_z$. It suffices to show that
$$
V = \bigcup_{z \in V} G_z.
$$
To see that $V \subseteq \cup_{z \in V} G_z$, let $z^* \in V$, and observe that $z^* \in G_{z^*} \subseteq \cup_{z \in V} G_z$.
To see that $V \supseteq \cup_{z \in V} G_z$, let $z \in V$, and observe that $z \in G_z \subseteq B_z \subseteq N$. Since $B_z \in \mathfrak{B}_z$, this shows that $z \in V$.
Q.E.D.
