Q:

Open surjection and filterbases problem

I have been struggling with the following problem. 
Let $f: X \to Y$ be an open surjection.  Prove that for each $x \in X$ and filterbase $\mathcal{B} \to f(x)$ there is a filterbase
  $\mathcal{U} \to x$ such that $f(\mathcal{U}) \vdash \mathcal{B}$ and $\mathcal{B} \vdash f(\mathcal{U})$
I tried the most straightforward filterbase $$\mathcal{U} = \{f^{-1}(B) \; | \; B \in \mathcal{B}\}$$
Does it works?  I was able to prove everything except convergence to $x$, but I wasn't able
to find a counterexample either.  If it works can you give me a tip on how to prove convergence? 
Any help or tips would be appreciated.  Thanks. 
Edit
A filterbase $\mathcal{B}$ is subordinated to $\mathcal{U}$ denoted by 
$\mathcal{B} \vdash \mathcal{U}$ if $\forall U \in \mathcal{U}$ exists $B \in \mathcal{B}$ such that $B \subseteq U$

A:

Claim: for every open set $O$ that contains $x$ and $B \in \mathcal{B}$, $f^{-1}[B] \cap O \neq \emptyset$.
Proof: $f[O]$ is an open neighbourhood of $f(x)$, as $f$ is open, so for some $B' \in \mathcal{B}$ we have $f[B'] \subset O$. As we have a filterbase, $B$ and $B'$ have a common refinement $B''$, with $B'' \subset B \cap B'$. Then for any $y \in B''$, $y \in B' \subset f[O]$, so $y = f(x)$ for some $x \in O$. As $f(x) \in B'' \subset B$, $x$ witnesses the required non-emptyness. 
Now $\mathcal{U} = \left\{ O \cap f^{-1}[B] : O \mbox{ open and contains } x, B \in \mathcal{B} \right\}$ I think will work as the required filterbase. 
Being a filterbase is easy to check (the non-emptyness of the elements was the claim).  For every $B$, $f^{-1}[B] \in \mathcal{U}$ and its image is $B$, which proves one subordination, and $f[ O \cap f^{-1}[B] ] \subset f[O] \cap B $, which contains $B$, which proves the other one.  Also, almost by definition $\mathcal{U} \rightarrow x$.

