= Solution
An <almost commutative algebra> over $\mathbb C$ has an exhaustive nonnegative increasing multiplicative <ascending filtration>
$$
0=F_{-1}R\subseteq F_0R\subseteq F_1R\subseteq\cdots,\qquad 1\in F_0R,\qquad F_iR\,F_jR\subseteq F_{i+j}R,
$$
for which the <associated graded ring>
$$
\operatorname{gr}R=\bigoplus_{n\geq0}F_nR/F_{n-1}R
$$
is a commutative finitely generated complex <algebra>. This includes the customary degree filtration with $F_0R=\mathbb C$, finitely many degree-one generators, and commutators lowering degree. The finite-generation clause is part of the convention used here and is necessary for the conclusion.
A finitely generated commutative complex <algebra> is a quotient of a <polynomial ring> $\mathbb C[X_1,\ldots,X_d]$. Repeated application of the central-variable case of the <Hilbert basis theorem> makes that polynomial <ring> <Noetherian>; quotients preserve the <ascending chain condition> by lifting ideal chains. Thus $\operatorname{gr}R$ is <Noetherian>.
We prove <ascending filtered-graded transfer of Noetherianity> directly. Give a <left ideal> $I\subseteq R$ the induced <ascending filtration> $I\cap F_nR$. Its <associated graded module> embeds into $\operatorname{gr}R$ as a homogeneous <ideal>, denoted $\operatorname{gr}I$. Choose finitely many homogeneous generators $\overline u_1,\ldots,\overline u_m$ of this ideal, of degrees $d_1,\ldots,d_m$, and lifts $u_j\in I\cap F_{d_j}R$. Homogeneous generators may be chosen by taking homogeneous components of any finite generating set.
For $u\in I\cap F_nR$, its degree-$n$ symbol can be expressed as
$$
\overline u=\sum_j\overline a_j\overline u_j,\qquad\deg\overline a_j=n-d_j,
$$
with terms of negative coefficient degree omitted. Lift each coefficient to $a_j\in F_{n-d_j}R$. Then
$$
u-\sum_ja_ju_j\in I\cap F_{n-1}R.
$$
Induction on $n$, ending at $F_{-1}R=0$, expresses $u$ in the <left ideal> generated by the $u_j$. Hence every <left ideal> is finitely generated. Applying the same argument to <right ideals>, or to the <opposite ring>, proves
$$
\boxed{R\text{ is left and right Noetherian}.}
$$
This degree-lowering proof terminates in finitely many steps and requires no completeness assumption on the <filtered ring>. Without finite generation of the graded algebra, a counterexample would be the commutative <polynomial ring> in infinitely many variables, whose ideals $(X_1)\subsetneq(X_1,X_2)\subsetneq\cdots$ do not stabilize.
For the <universal enveloping algebra> $U(\mathfrak g)$, use the degree <ascending filtration> inherited from the <tensor algebra> of the finite-dimensional <Lie algebra> $\mathfrak g$. For $x,y\in\mathfrak g$, the defining relation is $xy-yx=[x,y]\in F_1U(\mathfrak g)$, whereas the product $xy$ has filtration degree at most $2$. Thus the degree-one symbols commute in $\operatorname{gr}U(\mathfrak g)$. Those symbols generate the graded algebra, so it is a quotient of
$$
\operatorname{Sym}(\mathfrak g)\cong\mathbb C[X_1,\ldots,X_d],\qquad d=\dim_{\mathbb C}\mathfrak g.
$$
This already proves the required almost-commutative property. The <Poincaré-Birkhoff-Witt theorem> gives the stronger identification $\operatorname{gr}U(\mathfrak g)\cong\operatorname{Sym}(\mathfrak g)$, but injectivity of this map is not needed for the chain-condition argument. Therefore
$$
\boxed{U(\mathfrak g)\text{ is left and right Noetherian}.}
$$
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