A subset $A$ of a topological space $X$ is called projectively countable if for any continuous map $f:X\to\mathbb R$ the image $f(A)$ is countable. It is easy to see that each projectively countable set in a finite-dimensional cube $[0,1]^n$ is countable.
Problem. Is each projectively countable set in the Hilbert cube $[0,1]^\omega$ countable?
Added in Edit. The answer is affirmative under $\omega_1<\mathfrak c$ as shown by the following
Theorem 1. For any subset $X\subset K$ of cardinality $|X|<\mathfrak c$ in a compact metrizable space $K$ there exists a continuous function $f:K\to\mathbb R$ such that $f{\restriction}X$ is injective.
Proof. Consider the family $\mathcal H=\{H_{x,y}:x,y\in X,\;x\ne y\}$ of hyperplanes $H_{x,y}=\{f\in C(K):f(x)=f(y)\}$ in the Banach space $C(K)$. By the comment of @fedja to this question, the Banach space $C(K)$ cannot be covered by $<\mathfrak c$ hyperplanes. Consequently, there exists a function $f\in C(K)\setminus \bigcup \mathcal H$. This function has injective restriction $f{\restriction}X$. $\quad\square$
By a similar argument one can prove a linear version of the above theorem.
Theorem 2. For any subset $A\subset X$ of cardinality $|A|<\mathfrak c$ in a separable Banach space $X$ there exists a linear continuous functional $f:X\to\mathbb R$ such that the restriction $f{\restriction}A$ is injective.
Remark. Under CH the above problem has negative answer if this MO-problem has negative answer.
