Embed S in its fraction field and let E be a field extension that splits f*g. Every root of f*g is integral over C, hence contained in the integral closure of C in E. This means the coefficients of f (or g) lie in the integral closure of C in E. Yet they also lie in S, and since C is closed in S, they lie in C. Both f and g come from C[x].
This can be generalized to arbitrary commutative rings R and S. If every root of f*g is integral over C, the above reasoning holds. We only need show each root is integral over C.
Perhaps f*g already has some roots in S. If z is such a root, it is integral over C, and contained in S, and since C is the integral closure in S, z is in C. Divide f*g by x-z, and all the coefficients are still in C. Repeat this for all the roots that happen to lie in S, and call the quotient w(x).
Adjoin the indeterminant t to S, giving S[t], then divide by the ideal generated by w(t). Because w is monic, this ring, which I will call V, consists of polynomials in t, with coefficients in S, up to (but not including) the degreee of w. The coefficients of w lie in C, which lies in S, which lies in V. Think of w as a polynomial with coefficients in V.
Now w has at least one root in V, namely t. These roots are integral over C. Divide through by x-t, and do the same for any other roots. Find a new monic polynomial and start again.
Repeat the above procedure, adjoining new roots to build a tower of ring extensions. Each time, the new roots satisfy the original polynomial, and are integral over C. All the roots are integral over C, and that completes the proof.
Conversely, let f(x) lie in S[x], with f integral over R[x]. Now f satisfies a monic polynomial p(t), with coefficients in R[x].
Select k so that xk is larger than f. Let f′ = f - xk. Substitute f = f′+xk into p(t). Expand each term using the binomial theorem, and gather the terms divisible by f′ together. What remains is a monic polynomial in R[x], starting with xkn. Move this to the other side of the equation.
Thus f′ times something, which I will call g(x), lies in R[x]. Naturally f′*g lies in C[x]. Since k is large, f′ is monic. The product is also monic, hence g is monic. Invoke the lemma given above, and f′ lies in C[x], whence f lies in C[x].
In summary, the integral closure of R[x] in S[x] is C[x].
As a corollary, R integrally closed in S implies R[x] is integrally closed in S[x].
By induction, the same holds for finitely many indeterminants. for instance, R[x,y] is integrally closed in its fraction field F(x,y), and so on.
Let W be an infinite set of indeterminants, and consider the ring R[W] inside its fraction field F(W). Let z be a quotient of polynomials, exhibiting finitely many indeterminants from W. Assume z is integral over R[W]. A monic polynomial p makes this happen, and together, the coefficients of p exhibit finitely many indeterminants from W. Restrict attention to R adjoin finitely many indeterminants, covering p and z. Now z is integral over R adjoin x1 through xn, which is integrally closed, hence z lies in this base ring. Therefore an arbitrary polynomial extension of R remains integrally closed.