I'm trying to do a direct proof to show that Michael line is normal. Michael line is the topological space of $\mathbb{R}$ with the topology $\tau*=\{U \cup B : U \in \tau, B \subseteq \mathbb{R} - \mathbb{Q} \}$ Where $\tau$ Is the usual topology
Now let $F, K \subseteq \mathbb{R}$ be disjoint closed sets and we contruct for each $x \in F$ the sets $A_{x}$ as follows: if $x\notin \mathbb{Q}$ then $A_{x} = \{x\}$ Otherwise, since $F, K $ are disjoint then $F \subseteq \mathbb{R} - K$ and as $K$ is closed then, there exists $W \in \tau, B\subseteq \mathbb{R}-\mathbb{Q} $ such that $\mathbb{R} - K = W \cup B$ and as $x \in \mathbb{Q}$ It follows $x \in W$ and then $\exists \delta_{x} >0$ such that $B(x; \delta_{x}) \subseteq W$ in this case we choose $A_{x} = B(x; \delta_{x})$ obseve that every $A_{x}$ is open
Similarly we contruct for each $k \in K$ open sets $B_{k}$ as we did with every $A_{x}$
My claim is that $U = \bigcup_{x \in F} A_{x}, V = \bigcup_{k \in K} B_{k} \in \tau*$ are disjoint, I would appreciate any hint for that because I tried contradiction but didn't know how to continue the case in wich I have some $w \in B(x; \delta_{x}) \cap B(k; \delta_{k})$ with $x \in F, k \in K$
I suggest taking a slightly different approach. Let $F_0=F\cap\Bbb Q$ and $K_0=K\cap\Bbb Q$. Then $F_0$ and $K_0$ are closed in the topology that $\Bbb Q$ inherits from the Michael line, but that is simply its usual topology, so $F_0$ and $Q_0$ are disjoint closed sets in the metric space $\Bbb Q$. Thus, they have disjoint open nbhds $U_0$ and $V_0$ in $\Bbb Q$. For each $x\in F_0$ there is an $\epsilon_x>0$ such that $B_{\Bbb Q}(x,\epsilon_x)\subseteq U_0$ and $B_{\Bbb R}(x,\epsilon_x)\cap K=\varnothing$, where $B_{\Bbb R}$ denotes a ball in the usual metric on $\Bbb R$. Similarly, for each $x\in K_0$ there is an $\epsilon_x>0$ such that $B_{\Bbb Q}(x,\epsilon_x)\subseteq V_0$ and $B_{\Bbb R}(x,\epsilon_x)\cap F=\varnothing$. Let $U_1=\bigcup_{x\in F_0}B_{\Bbb R}(x,\epsilon_x)$ and $V_1=\bigcup_{x\in K_0}B_{\Bbb R}(x,\epsilon_x)$; clearly $U_1$ and $V_1$ are open in the Michael line, $U_1\cap V_1=\varnothing$, $F_0\subseteq U_1$, and $K_0\subseteq V_1$. Moreover, $U_1\cap K=V_1\cap F=\varnothing$, so $U=U_1\cup(F\setminus\Bbb Q)$ and $V=V_1\cup(K\setminus\Bbb Q)$ are disjoint open nbhds of $F$ and $K$ in the Michael line.