The problem:
Prove using Zorn's lemma that there exists a set $A \subseteq \mathbb{R}$ such that:
- for every non-zero polynomial $P(x_1,\dots, x_n)$ with rational coefficients and every $a_1, \dots a_n \in A$: $P(a_1, \dots, a_n) \neq 0$.
- for every $r \in \mathbb{R}$ there is a non-zero polynomial $P(x)$ with coefficients in $\mathbb{Q}(A)$ such that $P(r) = 0$
I've tried using Zorn's lemma by constructing a poset with the first condition on it (not sure if this is correct) and I 'hoped' it would also make the second condition true, but I'm not able to check this so I think I'm taking a wrong approach.
This is a fairly stragihtforward application of Zorn's lemma. Indeed, you want to consider the poset of all $A$'s with the first property. And it's easy to see that it satisfies Zorn's condition, since being an algebraically dependent set is witnessed by a finite subset.
Suppose now that $A$ is a maximal element given by Zorn's lemma, and $r\in\Bbb R$. If $P(r)\neq 0$ for all $P\in\Bbb Q(A)[x]$, then in particular for every finite $\{a_0,\ldots,a_{n-1}\}=A'\subseteq A$, $\Bbb Q(A',r)$ is a transcendental extension of $\Bbb Q(A')$, which means that every $P(x_0,\ldots,x_n)\in\Bbb Q[x_0,\ldots,x_n]$ it is true that $P(a_0,\ldots,a_n,r)\neq 0$.
But then $A\cup\{r\}$ also satisfies the condition, which is a contradiction to the maximality of $A$.