Equivalent condition with topological regular space by using local base. Let's have a definition of the 'regular space' as :

Let $(X,\tau)$ be topological space. If $\forall x\in X$,
  $\exists\mu_{X}$ : local base such that for $v\in\mu_{X}$, $v$ is
  closed.

Form here, I would like to prove an equivalent relation with our conventional
definition : 

Let $\left(X,\tau\right)$ be topological space.
Then $X$ is regular space if and only if for any $x\in X$ and $F$ :
  closed space which is not containing $x$, there exist $U$, $V\in\tau$
  such that $x\in U$, $F\subset V$ and $U\cap V=\phi$.

I'm struggling with 'Only if' condition.
For clarification, 

$\mu_{X}$ is called local base in $N(x)$ if $\forall v\in N(x)$,
  $\exists B\in\mu_{X}$ such that $x\in B\subset v$.

Moreover, the definition of $N(x)$ : 

$v\in N(x)$ if and only if $\exists O\in\tau$ such that $x\in O\subset v$.

 A: Suppose that $X$ is regular. It's more convenient to use a standard reformulation of regularity, which is close to what you want:
$X$ is regular iff 
(*) for every open set $O$ and every $x \in O$, there exists an open set $U$ such that $x \in U \subseteq \overline{U} \subseteq O$.
Left to right: For $x \in O$, $O$ open, $x \notin C:= X\setminus O$ and $C$ is closed.
So we have $U$, $V$ open and disjoint so that $x \in U$, $C \subset V$.
The latter means $(X \setminus O) \subset V$ so $X \setminus V \subseteq O$ and the former set is closed and contains $U$, so $\overline{U} \subseteq O$. So $U$ is as required.
Right to left is dual: suppose $x \notin C$ and $C$ closed. Then take $O = X\setminus C$ is open and contains $x$ so we have $U$ as promised. Then $x \in U$, $C \subseteq V:= X \setminus \overline{U}$ and $ U \cap V =\emptyset$.
Now, if $X$ is regular, and $O$ is open, take $U$ from the reformulation. Then $\overline{U}$ is a closed neighbourhood of $x$ (as witnessed by$U$) that sits inside $O$. So the closed neighbourhoods of $x$ form a local base at $x$.
And if the closed neighbourhoods form a local base, $X$ is regular: let $C$ be closed and $x \notin C$. Then $O := X\setminus C$ is open and contains $x$.
So let $D$ be a closed neighbourhood of $x$ with $D \subset X\setminus C$, this implies $C \subseteq V:= X\setminus D$. As $D$ is a neighbourhood of $x$ there is some open $U$ such that $x \in U \subseteq D$. Then $x \in U$, $C \subseteq V$ and $U \cap V \subseteq D \cap (X \setminus D) =\emptyset$ as required.
