Normal space on $[ -1,1]$ 
Consider the topology generated by the following  base  $$B = \{ [−1, b)\mid b > 0\} \cup \{(a, 1] \mid a < 0 \}$$ over $X = [−1, 1]$.  Is the space $X$ a normal space?

Edit:  my solution is the following
Let  $A$, $B$ be closed set such that $A\cap B = \emptyset$. The aim is to find two open sets of $X$ such that $A\subseteq U$, $B\subseteq V$, and $U\cap V = \emptyset$.
Since $B$ is closed then $\bar B= B$, so $A\cap\bar B=\emptyset$.
So $\forall a\in A$, $a\notin B$, so $\exists c < a$, if $a < 0$ then $(c, 1 ]\cap B=\emptyset$.
Then $U =\bigcup_{a\in A}(c, 1 ]$ is an open set and $A\subseteq U$
If $a >0$, $\exists c>a$ then $[-1 ,c)\cap B =\emptyset$.
Then $U =\bigcup_{a\in A}[-1, c)$ is an open set and $A\subseteq U$.
And the same thing for $B$. Is this right, and how I can show that $U\cap V = \emptyset$?
 A: 
Let  $A$, $B$ be closed set such that $A\cap B = \emptyset$. The aim is to find two open sets of $X$ such that $A\subseteq U$, $B\subseteq
V$, and $U\cap V = \emptyset$.

There is one problem here: $B$ is already defined to mean something else. You now have two different meanings for the symbol $B$ active at the same time. This is bad mathematical practice and I've seen many students end up hopelessly confused when they've done the same thing. There are 26 different letters, each with multiple fonts and two cases. There is no need to make some poor letter carry more than one burden of meaning at a time.

Since $B$ is closed then $\bar B= B$, so $A\cap\bar B=\emptyset$.
So $\forall a\in A$, $a\notin B$, so $\exists c < a$, if $a < 0$ then $(c, 1 ]\cap B=\emptyset$.

How does that follow?
For example, let $A = \{-1/2\}, B = [0,1/2]$. Then $A \cap B = \emptyset$, and there is an $a \in A$ with $a < 0$, but if $c < a$, then $(c, 1] \cap B \ne \emptyset$.
Yes, it turns out that neither $A$ nor $B$ is closed in this topology, so this is not a counter-example to $X$ being normal. But my point is that you need to explain why it must be that $(c,1] \cap B = \emptyset$.

how I can show that $U\cap V = \emptyset$?

You can't. The way you defined $U$ and $V$ makes no attempt at preventing them from overlapping.
And of course, Henno Brandsma has provided the counter-example to $X$ being normal. Every neighborhood of $\{-1\}$ must contain $0$, as must every neighborhood of $\{1\}$. And both of those sets are easily seen to be closed, as their complements are in the given sub-base for the topology.
