I found this proof in Kaplansky's Commutative Rings:

Induction on $\mbox{dim }A$.

$\mbox{dim }A =0 \ $:

Suppose $M\neq 0$. $\mbox{id}(M)=\mbox{depth}(A)=0$ so $M$ is injective, and hence is a direct sum of $\mbox{E}(A/\mathfrak{m})$. Since $A$ is Artin Gorenstein, $\mbox{E}(A/\mathfrak{m})\simeq A$ so $M$ is free.

$\mbox{dim }A \geq 1\ $:

$ 0\leftarrow M\leftarrow A^{n}\leftarrow K\leftarrow 0$

Then $\mbox{id}(K)<\infty$. Since $\mbox{dim }A\geq 1$ there is a non-zero-divisor $a\in A$, which is $K$-regular. Then $\mbox{id} _{A/(a)}(K/aK)\leq\mbox{id} _{A}(K)-1<\infty$ by one of the change of rings formulae. Now $A/(a)$ is Gorenstein and $\mbox{dim }A/(a)<\mbox{dim }A$ so by the induction hybothesis $\mbox{pd} _{A/(a)}(K/aK)<\infty$. By another change of rings formula, $\mbox{pd} _{A}(K)=\mbox{pd} _{A/(a)}(K/aK)<\infty $ so $\mbox{pd} _{A}(M)<\infty$ from the above short exact sequence.