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!
 A: The proof is wrong, as "$\varepsilon = \frac{d(a,b)}{2}\forall a \in a, \forall b \in B$" is ill-defined (the infimum of those numbers can be $0$ for disjoint closed sets).
Another, better idea is to note that $f_A: x \to d(x,A)$ is a continuous function from $X$ to $\mathbb{R}$ (where $d(x,A) = \inf \{ d(x,a): a \in A\}$ and which obeys $d(x,A) = 0$ iff $x \in \overline{A}$.
Then for disjoint closed $A$ and $B$, the function $$f(x) = \frac{d(x,A)}{d(x,A) + d(x,B)}$$ is well-defined (as no point $x$ exists such that $d(x,A) + d(x,B) = 0$, so that this term is always $>0$) and also continuous. Moreover $f[A] = \{0\}$ and $f[B]= \{1\}$ so that $U=f^{-1}[(-\infty,\frac13)]$ and $V = f^{-1}[(\frac{2}{3}, +\infty)]$ are disjoint open subsets of $A$ resp. $B$.
And $$A=(f_A)^{-1}[\{0\}] = \bigcap_n f^{-1}[(-\infty,\frac1n)]$$ 
shows that all closed sets are $G_\delta$'s and $(X,d)$ is even perfectly normal.
A: To give another proof more along the lines you started out with, using unions of open balls centered at the points of the sets:
Suppose we have $A,B$ disjoint and closed in a metric space $(X, d)$.
For each $a \in A$ we have that we have $r_a>0$ such that the open ball $B(a, r_a)$ is disjoint from $B$. This follows from $B$ being closed and $a \notin B$. So for these $r_A$ we have the property $$\forall a \in A : \forall b \in B: d(a,b) \ge r_a\tag{a}$$
Likewise for each $b \in B$ we have some $s_b>0$ such that $B(b,s_b) \cap A=\emptyset$ or
$$\forall b \in B: \forall a\in A: d(b,a) \ge s_b\tag{b}$$
Now define $O_A=\bigcup\{B(a,\frac{r_a}{2}): a \in A\}$ and $O_B=\bigcup\{B(b, \frac{s_b}{2}): b \in B\}$, which are both open as unions of (basic) open balls. Clearly $A \subseteq O_A$ and $B \subseteq O_B$.
Suppose that $p \in O_A \cap O_B$, which means that for some $a_0 \in A$ and some $b_0 \in B$ we have that $p \in B(a_0, \frac{r_{a_0}}{2})$ and $p \in B(b_0, \frac{s_{b_0}}{2})$. Define $r=\max(r_{a_0}, s_{b_0})$, and by $(a)$ and $(b)$ we have that $r \le d(a_0, b_0)$,
But then: $$r \le d(a_0, b_0) \le d(a_0,p)+ d(p, b_0) < \frac{r_{a_0}}{2} + \frac{s_{b_0}}{2} \le \frac{r}{2} + \frac{r}{2}=r$$
We now have the contradiction $r < r$ so the intersection of $O_A$ and $O_B$ is empty and we have shown normality.
A: 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.
