A space X is completely regular if and only if it is homeomorphic to a subspace of $[0,1]^J$ The following is from Munkres Topology: 

A space X is completely regular if and only if it is homeomorphic to a subspace of $[0,1]$$^J$ for some $J$. 

Note: $[0,1]$$^J$ is given the product topology.
I realize that $[0,1]$$^J$ is a completely regular space and any subspace of a completely regular space is also completely regular.  
However, why can we conclude for a space X that is homeomorphic to $[0,1]$$^J$ that it must also be completely regular?
I don't find the answer that homeomorphisms "preserve topological properties" convincing enough and am looking for something more specific. 
Thanks! 
 A: If $X$ is homeomorphic to $[0, 1]^{J}$, then there exists a homeomorphism
$$
\phi : X \rightarrow [0, 1]^{J}.
$$
To establish that $X$ is completely regular, we just need to verify the separation axioms $T_{1}$ and $T_{3 {1 \over 2}}$.  (For a detailed article, see Tychonoff space.
If $x \in X$, let $y = \phi(x)$.  Therefore, $\{x\} = \phi^{-1}(\{y\})$, i.e., the set $\{x\}$ is the pre-image of a closed set (namely, $\{y\}$) under the continuous mapping $\phi$, hence is closed in $X$.  This shows that the singletons in $X$ are closed; i.e., $X$ satisfies $T_{1}$.
Suppose $x \in X$, and suppose $C$ is a closed set in $X$ not containing $x$. To verify $T_{3 {1 \over 2}}$ for $X$, we need to show that there exists a continuous function $f : X \rightarrow \mathbb{R}$ such $f \equiv 1$ on $C$ and $f(x) = 0$.  (The function $f$ "separates" $x$ from $C$.)
To construct such an $f$, first note that
$$
y = \phi(x)
$$
is a singleton in $[0, 1]^J$, and the set $\phi(C)$ is closed in  $[0, 1]^J$.  Since  $[0, 1]^J$ satisfies $T_{3 {1 \over 2}}$, there exists a continuous function $g : [0, 1]^J \rightarrow \mathbb{R}$ such that
$$
g \equiv 1 \mbox{ on $\phi(C)$},
$$
and $g(y) = 0$.  Now take
$$
f = g \circ \phi.
$$
A: Any product of completely regular spaces is completely regular. So $[0,1]^J$ is completely regular for any $J$.
A subspace of a completely regular space is completely regular so any subspace of $[0,1]^J$ is completely regular.
If $X$ embeds into $[0,1]^J$, it is homeomorphic to a subspace of it, and a space homeomorphic to a completely regular space is completely regular. This is obvious from the definitions: we can move a point $x$ and a closed set not containing it by the homeomorphism to the completely regular space, get our real-valued function there and compose with the homeomorphism to get a separating function on the original space. 
So if $X$ embedds into $[0,1]^J$, it has to be completely regular. The reverse follows from the embedding theorem (separating points and point/closed sets, using all functions from $X$ to $[0,1]$, or those connected to a minimal base, say, get a smaller index set $J$).
