Proving every metrizable space is normal space 
A topological space $X,\tau$ is said to be normal space if for each pair of disjoint closed sets $A$ and $B$, there exists open sets $U$ and $V$ such that $A\subseteq U$,$B\subseteq V$ and $U\cap V=\emptyset$. Prove that every metrizable space is normal space.

If $X,\tau$ is a metrizable space then there exists $d:X\times X\to[0,+\infty]$ that defines the open sets in $\tau$.
Consider $\epsilon=\frac{d(a,b)}{2}\forall a\in A$ and $b\in B$.
Then $A\subset \bigcup_{a\in A} \mathscr{B}(a,\epsilon)$ since A is closed. The affirmation is proven in the following way:  some $a\in Fr(A)$ then $B(a,\epsilon)\cap Ext(A)\neq\emptyset$ 
In the same way $B\subset \bigcup_{b\in B} \mathscr{B}(b,\epsilon)$ 
If $U=\bigcup_{a\in A} \mathscr{B}(a,\epsilon)\\V=\bigcup_{b\in B} \mathscr{B}(b,\epsilon)$, 
then $U\cap V=\emptyset$.
Therefore $(X,d)$ is a normal space.
Question:
Is my proof right? If not. Why? 
Thanks in advance!

Let $X$ be a metrizable and let $d: X^{2} \rightarrow \mathbb{R} $ be a metric which defines the topology of $X$. We want to show that $X$ is Normal space. Let $ F_1$ and $F_1$ be disjoint closed subsets of $X$. Let
$$ (X-F_2) = \left\{ x\in X: d(x,F_1) < d(x,F_2) \right\}$$
and
$$ (X-F_1) = \left\{ x\in X: d(x,F_1) > d(x,F_2) \right\}$$
That is, since $F_1$ and $F_2$ are closed, then $(X-F_1)$ and $(X-F_2)$ are disjoint open sets in $X$ containing $F_2$ and $F_1$, respectively, i.e. $F_2 \subset (X-F_1)$ and $F_1 \subset (X-F_2)$. By definition of Normal space, hence $X$ is Normal.
