To extend the comment above.
The property: "$S\subset \mathbb R\setminus\mathbb Q$ is closed under addition and multiplication" satisfies Zorn's lemma, so there must be a maximal such set $S$.
Maximality means that for any irrational number $x\notin S$, there must be a polynomial of the form $p(z)-q$ where $q\in \mathbb Q$ and $p(z)$ is a non-zero non-constant polynomial with coefficients in $(S\cup\{0\})\oplus\mathbb N)$ such that $x$ is a root of the polynomial.
But if $S$ is countable, then the set of such polynomials is countable, and therefore the set of such roots is countable.
So any maximal $S$ must necessarily be uncountable.
This can be extended to a more general theorem: If $K\subset \mathbb R$ is uncountable and closed under addition and multiplication, then there is an uncountable $S\subseteq K\setminus\mathbb Q$ which is closed under addition and multiplication.
Proof: Since $K$ is uncountable, it contains a transcendental number, so the positive polynomials in that transcendental form a subset of $K\setminus\mathbb Q$ closed under addition and multiplication.
As before, by Zorn's lemma, there must be a maximal set $S\subset K\setminus\mathbb Q$ closed, and, as before, if $S$ is countable, then you get a contradiction - there are too many elements of $K\setminus (S\cup\mathbb Q)$ and not enough polynomials with coefficients in $(S\cup\{0\})\oplus\mathbb N$.