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.
