The set $\{f(X) \in \mathbb{R}[X] \hspace{1mm}|\hspace{3mm} f(n)\in \mathbb{Z} \hspace{3mm} \forall n \in \mathbb{Z}\}$ is countable. The set $\{f(X) \in \mathbb{R}[X] \hspace{1mm}|\hspace{3mm} f(n)\in \mathbb{Z}    \hspace{3mm} \forall n \in \mathbb{Z}\}$ is countable.
Clearly, $\mathbb{Z}[X]$ is contained in the set, from this previously asked question(({f(x) belongs to R[x] | f(n) belongs to Z for all n belongs to Z}uncountable or countable ? (TIFR GS 2022)) we have that all $\mathbb{Z}$ linear combinations of $$\binom{X}{m} = \frac{X(X-1)\cdots(X-m+1)}{m!}$$
I am able to prove that if $f(X)\in\mathbb{Q}[X]$ then $f(X)$ will be of the above form.
Now I am unable to prove what happens if at least one coefficient is irrational.Since sum of the coefficients is in $\mathbb{Z}$ we have at least two coefficients must be irrational.
suppose,
$f(X)=a_nX^n+a_{n-1}X^{n-1}+\cdot\cdot\cdot+a_0$
then since $f(\mathbb{Z}) \subset \mathbb{Z}$ , we have $a_0 \in\mathbb{Z}$ and $a_n+a_{n-1}+\cdot\cdot\cdot+a_0\in \mathbb{Z}$ $\implies$ $a_n+a_{n-1}+\cdot\cdot\cdot+a_1 \in \mathbb{Z}$
 A: It suffices to show that for every integer $d\ge 0$, the set
$$ S_d:=\{f\in\mathbb R[x]\mid \deg f \le d, f(\mathbb Z)\subset \mathbb Z\}$$
is countable, which we handle by induction.
The case $d=0$ is straightforward and serves as the basis for our induction. Inductively assume $d>0$. Let $f\in S_d$, and consider the first difference,
$(\Delta f)(x) = f(x+1)-f(x)$. Observe that leading terms cancel, so that $\Delta f \in S_{d-1}$. In other words, we have a map $$\Delta\colon S_d \to S_{d-1}$$ If $f$ and $g$ are mapped by $\Delta$ to the same polynomial, then for $h=f-g$, we have $\Delta(h)=0$, the zero polynomial. Note that $h$ is constant, because $h(x)-h(0)$ has infinitely many zeros (all $x\in \mathbb Z$). Meanwhile, $h(0) = f(0)-g(0)\in\mathbb Z$.
What we have proven is that there is an equivalence relation on $S_d$ ($f\sim g$ iff $\Delta f=\Delta g$) such that each equivalence class is countable, and inductively there are countably many equivalence classes. Therefore, $S_d$ is countable.
Exercise: Use a similar argument by induction above to prove the stronger claim that the elements $f$ of $S_d$ actually do have rational coefficients.
A: There is a much simpler way to prove this without explicitly determining what such a polynomial can be.  Namely, note that a polynomial of degree $d$ is uniquely determined by its values on $d+1$ points (because if $f$ and $g$ are two polynomials of degree $d$ that agree on $d+1$ points, then $f-g$ has degree at most $d$ but at least $d+1$ roots so it must be $0$).  So for each $d$, a polynomial $f$ of degree $d$ is determined by its values on $f(0),f(1),\dots,f(d)$.  If $f$ must map integers to integers, there are only countably many choices for these values, since each one must be an integer.  So there are only countably many such polynomials of degree $d$, and thus only countably many such polynomials of any degree.
