= Solution
Put $K=\operatorname{Frac}(A)$ and let $B$ be the <integral closure> of $A$ in $L$. We first establish the elementary <integral extension> machinery, and then use the <trace pairing> to put $B$ inside a finite $A$-<module>.
An <integral element> $x$ satisfies a <monic polynomial> over the base <ring>. If its equation has degree $r$, the powers $1,x,\ldots,x^{r-1}$ generate $A[x]$ as an $A$-<module>. More generally, adjoining finitely many <integral elements> gives a <finitely generated module>: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-<submodule> $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $xm_j=\sum_i a_{ij}m_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This <finite-module criterion for integrality> is the <determinant trick>.
If $x,y$ are <integral elements>, the finite $A$-<module> $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are <integral elements>. Thus the <integral closure> really is a subring. It also proves <integral dependence is transitive>: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many <integral elements> is a finite <integral extension>, even though an arbitrary <integral extension> need not be finite.
Choose a $K$-<basis> $v_1,\ldots,v_d$ of $L$. Each basis element is algebraic over $K$. If
$$
v_i^{r_i}+c_{i1}v_i^{r_i-1}+\cdots+c_{ir_i}=0,
$$
choose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_iv_i$ satisfies a monic equation with coefficients $a_i^jc_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-<basis>. This is <integral field basis by denominator clearing>.
We next prove the <trace of an integral element over a normal domain> property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are <integral elements> over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a <separable field extension>, that sum is the <field trace> $\operatorname{Tr}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a <normal domain>, equivalently an <integrally closed domain>, so
$$
\operatorname{Tr}_{L/K}(z)\in A\qquad(z\in B).
$$
The same conclusion applies to $ze_i$, since $B$ is a ring.
The <trace pairing> of a finite <separable field extension> is nondegenerate. One can see this directly using the allowed <Galois theory>: for a primitive element $u$, the embedding matrix of $1,u,\ldots,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of <basis>.
Let $M=\sum_iAe_i$. Its <trace-dual lattice> is
$$
M^\vee=\{x\in L:\operatorname{Tr}_{L/K}(xM)\subseteq A\}.
$$
Nondegeneracy supplies a trace-dual $K$-<basis> $e_1^*,\ldots,e_d^*$, with $\operatorname{Tr}(e_i e_j^*)=\delta_{ij}$. The coefficient formula gives
$$
M^\vee=\bigoplus_iAe_i^*,\qquad B\subseteq M^\vee.
$$
Equivalently, the matrix $T=(\operatorname{Tr}(e_ie_j))$ has entries in $A$ and nonzero determinant $D$; if $b=\sum_i c_i e_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-<module>.
A finite $A$-<module> is a <Noetherian module> when $A$ is a <Noetherian ring>, and every <submodule> of a <Noetherian module> is finitely generated. Since $B$ is an $A$-<submodule> of $M^\vee$, we obtain
$$
\boxed{B\text{ is a ring, integral over }A,\text{ and finite as an }A\text{-module}.}
$$
This proves <finiteness of integral closure in a finite separable extension>. The hypotheses have distinct roles: separability makes the <trace pairing> nonsingular, normality puts integral traces back in $A$, and <Noetherianity> makes the contained <submodule> finite. Also $\operatorname{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself <Noetherian> because its ideals are $A$-<submodules> of a finite $A$-<module>.
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