Countable subfield of $\mathbb{R}$ Let $f:\mathbb{R}→\mathbb{R}$ be a function.  Prove that there is a countable subfield $K⊂\mathbb{R}$such that $f(K)⊂K$.
What I've tried: start with any $x$, look at the set ${x, f(x), f(f(x)), ...}$ Clearly it satisfies $f(K)⊂K$, but I can't prove it's a subfield. Probably it's a false start.
Then I thought maybe looking at $\mathbb{Q}$, but don't know how to continue.
Would appreciate if someone can give me a hint first, instead of a full solution.

Let $K_0=\mathbb Q$ and, for each $n\in\mathbb N$, let $K_n$ be the smallest subfield of $\mathbb R$ containing $K_{n-1}\cup f(K_{n-1})$. Finally, let $K=\bigcup_{n=0}^\infty K_n$. Then $K$ is a countable subfield of $\mathbb R$ and $f(K)\subset K$. Can you prove that?
