Regarding Compact Semigroup I have a problem in understanding the proof of following Theorem 2 (b), which is given in the book "The Theory of Topological Semigroups" by J. H. Carruth, J. A. Hildebrant and R. J. Koch as Theorem 1.42 at page 34.
Let $S$ be a semigroup and $\mathcal{U}$ is a non-empty subset of $S \times S$, then $\mathcal{U}$ is said to be a $\Delta$-ideal if $\Delta \mathcal{U} \cup \mathcal{U} \Delta \subset \mathcal{U}$, where the multiplication in $S \times S$ is coordinate-wise multiplication and $\Delta=\{(x,x)\mid x \in S\}$. 
A topological semigroup $S$ is a Hausdorff topological space such that the binary operation is continuous.
Theorem 1: Let $X, Y$ and $Z$ be topological spaces, $A$ a compact subset of $X$, $B$ a compact subset of $Y$, $f:X \times Y \rightarrow Z$ be a continuous function, and $W$ an open subset of $Z$ containing $f(A \times B)$. Then there exists and open set $U$ in $X$ and an open set $V$ in $Y$ such that $A \subset U$, $B \subset V$ and $f(U \times V) \subset W$.
Theorem 2: Let $S$ be a semigroup and $\mathcal{V}$ be a subset of $S \times S$ containing $\Delta$. Then $\mathcal{U}=\cup\{\mathcal{W} \mid \mathcal{W} \text{ is a } \Delta-\text{ideal and } \mathcal{W} \subset \mathcal{V}\}$ is a $\Delta$-ideal. Moreover
$(a)$ If $S$ is a topological semigroup and $\mathcal{V}$ is closed, then $\mathcal{U}$ is closed and
$(b)$ If $S$ is a compact semigroup and $\mathcal{V}$ is open, then $\mathcal{U}$ is open.
Proof: $(b)$ Suppose that $S$ is a compact semigroup and $\mathcal{V}$ is open. Fix $(x,y) \in \mathcal{U}$. Then $\{(x,y)\} \cup (x,y) \Delta \cup \Delta (x,y) \cup \Delta (x,y) \Delta \subset \mathcal{V}$ and hence by repeated application of Theorem 1, there is an open subset set $V$ containing $(x,y)$ such that $V \cup V \Delta \cup \Delta V \cup \Delta V \Delta \subset \mathcal{V}$. This set is obviously a $\Delta$-ideal and hence $(x,y) \in V \subset \mathcal{U}$.
I do not understand, how one can obtain an open subset $V$ by repeated application of Theorem 1.
 A: Let $\mathcal{V}$ be an open subset of $S \times S$ and let $(x,y) \in \mathcal{U}$. Since $S$ is Hausdorff, the singleton $\{(x,y)\}$ and the diagonal $\Delta$ are closed and hence compact in $S \times S$. 
Let $f: (S \times S) \times (S \times S) \to (S \times S)$ be the map defined by $f((a_1,b_1),(a_2,b_2)) = (a_1a_2, b_1b_2)$. Since $S$ is a topological semigroup, $f$ is continuous. 
Applying Theorem 1 to $f$ with $X = Y = Z = S \times S$, $A = \Delta$, $B = \{(x,y)\}$ and $W = \mathcal{U}$, there exists open subsets $U$ and $V_0$ of $S \times S$ such that $\Delta \subset U$, $(x,y) \in V_0$ and $f(U \times V_0) \subset W$, and thus $\Delta V_0 \subset UV_0 \subset \mathcal{U}$.
A similar argument would show that there exists open subsets $V_1$ and $V_2$ containing $(x,y)$ such that $V_1\Delta \subset \mathcal{U}$ and $\Delta V_2\Delta \subset \mathcal{U}$. Now the open set $V = \mathcal{U} \cap V \cap V_1 \cap V_2$ satisfies $(x,y) \in V$ and $V \cup V \Delta \cup \Delta V \cup \Delta V \Delta \subset \mathcal{V}$.
