= Solution
{c}
The zero means and orthogonality give the exact second moment
$$
\mathbb E A_N^2=\frac1{N^2}\sum_{n,m\leq N}\mathbb E(X_nX_m)=\frac1N.
$$
Thus $A_N\to0$ in $L^2$. <Chebyshev's inequality> gives the requested <weak law of large numbers>:
$$
\boxed{\mathbb P(|A_N|\geq\epsilon)\leq\frac1{N\epsilon^2}\longrightarrow0.}
$$
Independence is unnecessary; the stated second-moment orthogonality is sufficient.
For an explicit almost surely convergent subsequence take $N_j=2^j$. For each positive integer $m$,
$$
\sum_j\mathbb P(|A_{2^j}|\geq1/m)\leq m^2\sum_j2^{-j}<\infty.
$$
The <First Borel-Cantelli lemma> makes only finitely many such exceedances occur. Intersecting the resulting probability-one events over $m$ gives \b[$A_{2^j}\to0$ almost surely].
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