= Solution
{c}
\b[No.] Take $R=k[t]$ over a <field> $k$, and inside $R[x]=k[t,x]$ take
$$
S=k[t]+tx\,k[t,x].
$$
This is a subring containing $R$: its elements have arbitrary constant coefficient in $k[t]$ as polynomials in $x$, and every positive-degree coefficient is divisible by $t$; that condition is preserved by sums and products. The <ring> $R$ is <Noetherian>. Indeed, a nonzero <ideal> of $k[t]$ has an element $g$ of least degree, and division by $g$ shows that every element of the <ideal> is a multiple of $g$. Thus $k[t]$ is a <principal ideal domain>, and its <ideals> are finitely generated.
In $S$, let $J_n=(tx,tx^2,\ldots,tx^n)_S$. These form an ascending chain. We show that $tx^{n+1}\notin J_n$. If it belonged, we could write
$$
tx^{n+1}=\sum_{j=1}^n s_jtx^j,\qquad s_j\in S.
$$
Cancel $t$ in the <integral domain> $k[t,x]$, then apply the <evaluation homomorphism> $t\mapsto0$. Every $s_j\in S$ becomes a scalar in $k$, so this would give
$$
x^{n+1}=\sum_{j=1}^n s_j(0,x)x^j,\qquad s_j(0,x)\in k,
$$
which is impossible by comparison of degrees. Hence $J_n\subsetneq J_{n+1}$ for every $n$, and
$$
\boxed{k[t]\subseteq k[t]+txk[t,x]\subseteq k[t,x],\quad
k[t]\text{ Noetherian but }k[t]+txk[t,x]\text{ not Noetherian}.}
$$
This is a <constant-plus-ideal non-Noetherian subring> construction: the restricted coefficients prevent a finite list of positive powers of $x$ from generating all the required powers.
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