Here is a sketch of a proof.
Let A_0 be the set of all roots of polynomial equations in one variable with integer coefficients.
(1) A_0 includes all rational numbers.
If m, n in ZZ with n!=0 then nx-m = 0 has root x = m/n.
(2) A_0 is closed under negation.
Suppose x_1 is a root of:
a_n x^n + a_(n-1) x^(n-1) + ... + a_0 = 0.
Then -x_1 is a root of:
(-a_n)^n x^n + (-a_(n-1))^(n-1) x^(n-1) + ... + a_0 = 0.
(3) A_0 is closed under reciprocal.
Suppose x_1 != 0 is a root of:
a_n x^n + a_(n-1) x^(n-1) + ... + a_0 = 0
Then 1/x_1 is a root of:
a_0 x^n + a_1 x^(n-1) + ... + a_n = 0
(4) A_0 can equivalently be described as the set of all roots of monic polynomial equations in one variable with rational coefficients.
Given a polynomial equation with integer coefficients:
a_n x^n + a_(n-1) x^(n-1) + ... + a_0 = 0
divide all the coefficients by a_n to get a monic polynomial with rational coefficients.
Conversely, given a monic polynomial equation with rational coefficients, multiply all of the coefficients by the least common multiple of the denominators of the coefficients to get a polynomial equation with integer coefficients.
(5) The coefficients of a monic polynomial of degree n are the n elementary symmetric polynomials in the n zeros
Proof omitted.
(6) A_0 is closed under addition.
Given polynomial equations:
P(x) = x^n + a_(n-1) x^(n-1) + ... + a_0 = 0
Q(x) = x^m + b_(m-1) x^(m-1) + ... + b_0 = 0
where a_i and b_j are rational.
Denote the n roots of P(x) by p_1,...,p_n and the m roots of Q(x) by q_1,...,q_m.
Let R(x) = prod_(i=1)^n prod_(j=1)^m (x-(p_i+q_j))
Then R(x) is a monic polynomial, each of whose coefficients is symmetric in p_i and q_j, so it is expressible in terms of integer multiples of the elementary symmetric polynomials in p_i and q_j, namely the coefficients a_i and b_j of P(x) and Q(x). So the coefficients of R(x) are rational too. So the zeros of R(x) are in A_0.
(7) A_0 is closed under multiplication
Given polynomial equations:
P(x) = x^n + a_(n-1) x^(n-1) + ... + a_0 = 0
Q(x) = x^m + b_(m-1) x^(m-1) + ... + b_0 = 0
where a_i and b_j are rational.
Denote the n roots of P(x) by p_1,...,p_n and the m roots of Q(x) by q_1,...,q_m.
Let R(x) = prod_(i=1)^n prod_(j=1)^m (x-p_i q_j)
Then R(x) is a monic polynomial, each of whose coefficients is symmetric in p_i and q_j, so it is expressible in terms of integer multiples of the elementary symmetric polynomials in p_i and q_j, namely the coefficients a_i and b_j of P(x) and Q(x). So the coefficients of R(x) are rational too. So the zeros of R(x) are in A_0.
(8) A_0 is algebraically closed
Suppose P(x) = x^n + a_(n-1) x^(n-1) + ... + a_0
where a_i in A_0 for i = 0,...,n-1.
For each i in 0,...,n-1 let P_i(x) be a monic polynomial with rational coefficients of minimal degree n_i such that P_i(a_i) = 0. Denote the n_i zeros of P_i(x) = 0 as a_(i,1), ... , a_(i,n_i).
Consider:
R(x) = prod_(j_(n-1) = 1)^(n_(n-1)) prod_(j_(n-2) = 1)^(n_(n-2)) ... prod_(j_0 = 1)^(n_0) (x^n + a_((n-1),j_(n-1)) x^(n-1) + a_((n-2),j_(n-2)) x^(n-2) + ... + a_(0,j_0))
That is, R(x) is the product of all possible variants of P(x), substituting the other roots of the minimal polynomial equations of each coefficient.
Then each of the coefficients of R(x) is symmetric in all of the roots of each of the minimal monic polynomials, so is expressible in terms of integer multiples of the elementary symmetric polynomials, which are the rational coefficients of each of the P_i(x) polynomials.
So the coefficients of R(x) are all rational and its zeros include the zeros of P(x). So the zeros of P(x) are in A_0 too.
