= Solution
To obtain a <countable closure under real polynomial roots>, start with $B_0=\{1,\pi\}$ and explicitly retain previous elements at each stage:
$$
B_{n+1}=B_n\cup\phi(B_n),\qquad X=\bigcup_{n=0}^{\infty}B_n.
$$
Part (b) shows that $\phi$ takes <countable sets> to <countable sets>; consequently all $B_n$ and then $X$ are countable by the <countable union of countable sets> theorem. Also $1,\pi\in X$.
For a nonzero <polynomial> $P(t)=c_0+\cdots+c_dt^d$ with all $c_j\in X$, choose stages $n_j$ containing the finitely many coefficients and put $N=\max_jn_j$. The inclusions $B_n\subseteq B_{n+1}$ ensure that every coefficient belongs to $B_N$. Every real <root of a polynomial> $P$ therefore belongs to $\phi(B_N)\subseteq B_{N+1}\subseteq X$. \b[This <countable set> $X$ has all the required properties.] Retaining $B_n$ explicitly avoids assuming the generally false inclusion $A\subseteq\phi(A)$: for example, $\phi(\{1\})$ does not contain $1$, since a nonzero <polynomial> all of whose coefficients are one is positive at $1$.
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