Verifying separation axioms for the product topology $\{0,1\}^J$ Is my proof efficient? I think I was able to verify the separation axioms but I am still not entirely sure. Any help is greatly appreciated! Thanks!
$\def\R{{\mathbb R}}$
I wish to prove the following: If $(X,T)$ is zero-dimensional and completely regular, then there is a set $J$ such that $X$ is homeomorphic to a subspace of $\{0,1\}^J$.
$\textbf{Solution:}$ If $X$ is homeomorphism to $\{0,1\}^J$ then there exists a homeomorphism $$\varphi\colon X\to \{0,1\}^J$$ where $X$ is completely regular. We need to verify the separation axioms $T_1$ and $T_{3^{1/2}}$. 
Let $x\in X$ and let $y=\varphi(x)$. Then $\{x\} = \varphi^{-1}\{y\}$, i.e. the set $\{x\}$ is the pre-image of a closed set $\{y\}$. By the continuous mapping $\varphi$ is closed in $X$, singletons in $X$ are closed, and $X$ satisfies $T_1$. 
Now, let $x\in X$ and suppose $C$ is a closed set in $X$ not containing $x$. To verify $T_{3^{1/2}}$ for $X$, we must show there exists a continuous functions $f\colon X\to \R$ such that $f$ is homeomorphic on $C,$ $f(x)= 0.$ To construct such an $f$, $y=\varphi(x)$ is singleton in $\{0,1\}^J$ and the set $\varphi(C)$ is closed in $\{0,1\}^J$. Since $\{0,1\}^J$ satisfies $T_{3^{1/2}}$, there exists a continuous function $g\colon \{0,1\}^J \to \R$ and $g$ is homeomorphic on $\varphi(C)$ so $g(y) = 0$ and $f=g \circ \varphi$. Thus, $X$ embedds into $\{0,1\}^J$. Thus, is completely regular using all functions from $X$ to $\{0,1\}.$
 A: You can deduce the result from a very general embedding theorem (proved in Engelking, Munkres, Willard, etc and many other text books):
Suppose $X$ is a space and $\mathcal{F} = \{f_i: X \to Y_i \mid i \in I\}$ is a family of continuous functions from $X$ to spaces $Y_i$ that obeys two conditions:


*

*$\mathcal{F}$ separates points: for every pair of distinct points $x,y$ of $X$, there is some $f_i \in \mathcal{F}$ such that $f_i(x) \neq f_i(y)$.

*$\mathcal{F}$ separates points and closed sets: whenever $C$ is closed in $X$ and $p \notin C$, there exists some $f_i \in \mathcal{F}$ such that $f_i(x) \notin \overline{f_i[C]}$ (closure in $Y_i$ of course).
Then $e:X \to Y:=\prod_{i \in I} Y_i$ defined by $(e(x))_i = f_i(x)$ is an embedding from $X$ into $Y$ (so $e: X \simeq e[X]$).
This theorem is the "reason" we an embed all $T_{3 \frac12}$ spaces into Tychonoff cubes $[0,1]^I$, e.g.
The crux of this exercise to find such a family of functions on $X$ and check the aforementioned conditions for them. 
Hint: let $\{B_i\mid i \in I\}$ be a base for $X$ of clopen sets (by zero-dimensionality) and let $f_i$ be the characteristic function of $B_i$ from $X$ to $\{0,1\}$. This corresponds to Brian's hint in the comments.
