You can prove it with Stein's method, however it's debatable if the proof is elementary. The plus side of Stein's method is you get a slightly weaker form of Berry Esseen bounds essentially for free. Also, Stein's method is nothing short of black magic! You can find an exposition of the proof in section 6 of this link. You'll find other proofs of the CLT in the link as well.
Here's a brief outline:
1) Prove, using simple integration by parts and the normal distribution density, that $Ef'(A)-Xf(A)=0$ for all continuously differentiable iff $A$ is $N(0,1)$ distributed. It's easier to show $A$ normal implies the result and a bit harder to show the converse, but perhaps it can be taken on faith.
2) More generally, if $Ef(X_n)-X_nf(X_n)\rightarrow 0$ for every continuously differentiable $f$ with $f,f'$ bounded, then $X_n$ converges to $N(0,1)$ in distribution. The proof here is again by integration by parts, with some tricks. Specifically, we need to know that convergence in distribution is equivalent to $Eg(X_n)\rightarrow E g(A)$ for all bounded continuous functions $g$. Fixing $g$, this is used to reformulate:
$$Eg(X_n)-Eg(A)=Ef'(X_n)-X_nf(X_n),$$
where one solves for $f$ using basic ODE theory, and then shows $f$ is nice. Thus if we can find such a nice $f$, by assumption the r.h.s. goes to 0, and therefore so does the left side.
3) Finally, prove the central limit theorem for $Y_n:=\frac{X_1+\cdots+X_n}{\sqrt{n}}$ where $X_i$ are iid with mean 0 and variance 1. This again exploits the trick in step 2, where for every $g$ we find an $f$ such that:
$$Eg(X_n)-Eg(A)=Ef'(X_n)-X_nf(X_n).$$