My attempt:
Let $X_1,X_2$ be two completely regular spaces and $W$ their product. Let $a=(a_1,a_2) \in W$ and $B \subset W$ be a closed set disjoint from $a$. Choose basis element $U=(U_1,U_2)$ disjoint of $B$. By complete regularity of $X_1,X_2$, choose $f_i: X_i \rightarrow [0,1]$ such that $f_i (a_i)=1, f_i (X \setminus U_i)={0}$. Define $g_i (x)=f_i (p_i (x))$ where $x \in W$ and $p_i$ is a projection. Then $f(x)=g_1(x)g_2(x)$ is our desired function for complete regularity.
Does this seem correct?
Best Answer
Yes, your proof is correct. However, we can even say more: The product $X=\prod_i X_i$ of a arbitrary family of completely regular spaces $(X_i)_i$ is completely regular. This can be proven by a generalization of your approach. Another approach is via initial topologies. Note that a space is completely regular if and only if it has the initial topology with respect to some family of real-valued continuous maps. Since the product $X$ has the initial topology with respect to all projections $p_i:X\to X_i$, it has the initial topology with respect to a family of continuous real-valued maps $X\to X_i \to \Bbb R$ (see for example the Transitive law of initial topologies near the end of Henno Brandsma's answer here), and is thus completely regular.