A regular, Lindelöf space is normal Following a reference from "General Topology" by Stephen Willard

Well since the difference between an open set and a closed set is an open set and since the union of open sets is an open set, it is clearly that $S$ and $T$ are open; but unfortunately I can't prove that $S$ and $T$ are disjoint open set containing $A$ and $B$. Anyway it seems to me that $U=\bigcup_{i=1}^nU_i$ and $V=\bigcup_{i=1}^nV_i$ are two disjoint open sets that containing $A$ and $B$ and so I ask: why define the set $S$ and $T$? Could someone help me?

Hint: If they are not disjoint, then it is necessary that for some $m$ and $n$, $S_m$ and $T_n$ are also not disjoint. Show that this is not possible by considering two cases: first, when $m>n$, and the other, when $m\leq n$.  
Further Explanation: 

 Working off the Hint, suppose $\exists$ some $x\in S_m\cap T_n$ for some $m$ and $n$ s.t. $m> n$. By the definition of $S_m$, we have that $x\in U_m-\overline{\left(\cup_{j=1}^{m-1} T_j\right)}$. But this means $x\not\in \overline{T_j}$ for any $j\in {1,2,\dots,m-1}$. Since we assumed $m>n$, this means in particular that $x\not\in \overline{T_n}$ so $x\not\in T_n$, contradicting our initial assumption. The other case may be done in a similar fashion.  

EDIT: 
Hint 2: Each $a\in A$ lies in some $U_i$. At the same time, each $T_j$ lies in $V_j$. Is it possible for $a$ to be contained in any $T_j$? What about any $\overline{T_j}$? Then where must $a$ lie? 
Additional Explanation: 

 Since each $T_j$ lies in $V_j$, and, for each $j$, $\overline{V_j}\cap A=\emptyset$, $a\not\in \overline{V_j}$ and so, $a\not \in \overline{T_j}$ for any $j$. In particular, this means $a\not\in \cup_{j=1}^{i-1}\overline{T_j}=\overline{\cup_{j=1}^{i-1} T_j}$, so $a\in U_i-\overline{\cup_{j=1}^{i-1} T_j}=S_i$. $a$ was an arbitrary element of $A$, so $A\subset \cup_{k=1}^{\infty} S_k$. Reason similarly for $B$.

