Proof Verification: Compact Hausdorff Implies Normal

Question

Here is the definition of "normal space" that I am working with:

A topological space $(X,\tau)$ is normal if for any two disjoint closed sets $A$ and $B$, there exists disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $B \subseteq V$.

Prove that every compact Hausdorff space is a normal space.

Here is my attempt at a proof:

1. Suppose $(X,\tau)$ is a compact Hausdorff space. Let $A$ and $B$ be disjoint closed subsets in $X$. Since closed subsets of Hausdorff and compact spaces are Hausdorff and compact, $A$ and $B$ are also compact Hausdorff spaces.
2. Since $A$ and $B$ are compact, every open covering has a finite subcovering. Thus $A \subseteq \displaystyle{\Pi_{i\in I}U_i}$ has finite subcovering $A \subseteq U_1 \cup ... \cup U_n$ and $B \subseteq \displaystyle{\Pi_{j\in J}V_j}$ has finite subcovering $B \subseteq V_1 \cup ... \cup V_m$, where $U_i$ and $V_j$ are open sets for any $i \in I$, $j \in J$.
3. Let $U=U_1 \cup ... \cup U_n$ and $V=V_1 \cup ... \cup V_m$. Then, $U \cap V = (U_1 \cup ... \cup U_n) \cap (V_1 \cup ... \cup V_m)= (U_1 \cap V_1) \cup ... \cup (U_n \cap V_m) = \phi$ since $U_i \cap V_j = \phi$.

Thus $U$ and $V$ are disjoint open sets, and $(X,\tau)$ is normal.

There are a few doubts that I have concerning this proof:
Firstly, is it true that $U_i \cap V_j = \phi$ ? I think it is because $(X,\tau)$ is Hausdorff, so open neighbourhoods about distinct points are separated.
Secondly, did I distribute the intersection correctly in step 3?

Answer 1

Be carefull: In $3$ are more terms and isn't true $U \cap V$ will be no necesary empty.

One way is a follow:

As you say, every cover by open sets $A \subseteq \cup_{i\in I}U_i$ has a finite covering, i.e. $A \subseteq U_1 \cup \ldots \cup U_n$.

Then, for $U_1$, will be find a finite open cover of $B$ that is disjoint:

As $X$ its Hausdorff, for every $y \in B$ we can find a open set $V_y^1$ such that $y \in V_y^1$ and $U_1 \cap V_y^1 = \emptyset$. We can do this for every $y \in B$. It's clear that $B \subseteq \cup_{j\in J}V_j^1$. As $B$ it's Hausdorff (as you say) then we can extract a finite subset, i.e. $B \subset V_{j_1}^1 \cup \ldots \cup V_{j_n}^1:= \tilde{V}^1$. It's clear that $V_{j_1}^1 \cup \ldots \cup V_{j_n}^1$ and $U_1$ are disjoint by construction.

We can repeat this for every open set $U_2, \ldots, U_n$ of the finite covering of $A$. At the end we have $n$ finite coverings of $B$, called $\tilde{V}^1, \ldots, \tilde{V}^n$, and having the property that $\tilde{V}^j \cap U_j = \emptyset$ for all $j$.

Finally take $W = \tilde{V}^1 \cap \ldots \cap \tilde{V}^n$. As every $\tilde{V}^j$ is open, $W$ is open. As $B \subset \tilde{V}^j$ for every $j$ (this is by construction), $B \subset W$. Finally as $\tilde{V}^j \cap U_j = \emptyset$ for every $j$, then $W \cap (U_1 \cup \ldots \cup U_n) = \emptyset$.

If we define: $U= U_1 \cup \ldots \cup U_n$, then $U$ and $W$ are the disjoint open sets we were looking for.