Neighborhood of Diagonal induces Open Cover

Theorem
Let $T = \struct{X, \tau}$ be a Topological Space.

Let $X \times X$ denote the cartesian product of $X$ with itself.

Let $\tau_{X \times X}$ denote the product topology on $X \times X$.

Let $T \times T$ denote the product space $\struct {X \times X, \tau_{X \times X} }$.

Let $U$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in $T \times T$.

Let $\VV = \set{V : V \times V \subseteq U}$

Then:
 * $\VV$ is an open cover of $T$

Proof
Let $x \in X$.

By definition of neighborhood:
 * $\exists W \in \tau_{X \times X} : \tuple{x, x} \in W : W \subseteq U$

By definition of product topology:
 * $\BB = \set {W_1 \times W_2: W_1, W_2 \in \tau}$ is a basis on $T \times T$

By definition of basis:
 * $\exists W_1, W_2 \in \tau : \tuple{x, x} \in W_1 \times W_2 : W_1 \times W_2 \subseteq W$

From Subset Relation is Transitive:
 * $\exists W_1, W_2 \in \tau : \tuple{x, x} \in W_1 \times W_2 : W_1 \times W_2 \subseteq U$