I recently answered a question, in which I proved that If a polynomial fixes the unit circle then $P$ is a monomial (a classical result),i,e: $$\forall P\in \Bbb C[X]\ \ \ \ (\forall z\in \Bbb C \ \ |z|=1\implies |P(z)|=1)\implies P=aX^n \tag 1$$
Let $U$ denote the unit circle, this result is equivalent to: the only polynomials $P$ such that $P(U)\subset U$ are monomials.
My question: Given an infinite countable subset $A$ of $U$ is it true that if a polynomial $P$ verifies $P(A)\subset U$ then $P$ is a monomial.
My attempt
I replaced $A$ countable in the question by $U\backslash A$ countable, and I proved using the same method as $(1)$ is the monimials, e.g: if a polynomial $|P(z)|=1$ for all complex numbers $z$ such that $|z|=1$ and $z\not\in A$ for a countable subset of $U$ then this polynomial can be written as $P=aX^n$.
For the question I tried to find a polynomial which is not monomial and verify all conditions but without success,
Edit (As stated in the comments by LeGrandDoDom) when $A$ is dense in the unit circle then we have $|P(z)|$ is continue therefore, this brings us to the case $(1)$ which is proved. The remaining case is when $U$ is a countable no dense subset of $U$.
In particular we have: if $|P(z)|=1$ for evry root of unity $z$ then $P=aX^n$ for some integer $n$ and a constant $a\in U$.
Forget complex analysis, this is a question of algeraic geometry:
Let $A\subset U$ be infinite and $P$ a polynomial with $P(A)\subset U$. Rewrite in the reals, $P(z)=Q(x,y)+iR(x,y)$ with $Q,R\in\mathbb R[x,y]$. Then $f(x,y):=Q(x,y)^2+R(x,y)^2-1$ vanishes for infinitely many points on the curve $X^2+Y^2-1=0$. (Now we go back to the complex number field because we need an algebraically closed field.) Hence the ideal $I:=(f, X^2+Y^2-1)\subseteq \mathbb C[X,Y]$ has an infinite vanishing set $V(I)\subset \mathbb C^2$, which is the union of finitely many varieties (irreducible components). These are all contained in the variety(!) $V(X^2+Y^2-1)$. As this is one-dimensional, all proper subvarieties are finite sets. So $|V(I)|=\infty$ implies that in fact $V(I)=V(X^2+Y^2-1)$. This means that $f$ vanishes on all of $U$, hence $P(U)\subseteq U$ and we can use your first result to complete the proof.