Continuous functions and restrictions to a countable domain Let $I$ a compact set and $f\in\mathcal{C}([0,1]^I)$.
Then exists $J\subset I$, countable or finite, such that:
if $x,y\in [0,1]^I$ such that $x\big|_J=y\big|_J\Rightarrow f(x)=f(y)$.
 A: The result is true for the set of all functions from $I$ into $[0,1]$ (not only the continuous ones). 
Claim. For any open subset $U\subset [0,1]^I$ there exists $J\subset I$ countable so that $x\in \overline U$ and $x|_J=y|_J$ imply $y\in \overline U$.
Find a maximal pairwise disjoint family of standard basic nonempty open subsets of $U$, by the Hewitt-Marczewski-Pondiczery theorem this family is countable, let us call it $(U_n)$. For each $n$, let $J_n$ be the set of non-trivial coordinates in the basic open set $U_n$. It should be clear that $J=\bigcup_n J_n$ satisfies the requirements of the claim.    
Let $\{ V_n:n\in\mathbb N\}$ be a countable base of $\mathbb R$. For each n, we can find a countable subset $J_n$ of $I$ satisfying the conclusion of the Claim with respect to the open set $f^{-1}(V_n)$. Now, we claim that   $J=\bigcup J_n$ is the required set. Suppose $x|_J=y|_J$ and $f(x)\ne f(y)$ then there are   disjoint open sets $V_n, V_m$ so that $f(x)\in V_n$ and $f(y)\in V_m$. This implies that $x\in f^{-1}(V_n)=U_n$ and $f^{-1}(V_m)=U_m$,  the sets $U_n, U_m$ are disjoint. But by the claim $y\in U_m$ and $x|_{J_m}=y|_{J_m}$ implies $x\in \overline{U_m}$ which is impossible.
