= Solution
Expand $X_tX_{t'}$ into products of the independent mean-zero signs. Each increment is the sum of two monomials, $a_t$ and $a_{t-2}a_{t-1}$. At distinct times, no monomial from one increment has exactly the same set of sign indices as a monomial from the other. Every product therefore contains at least one sign to an odd power, whose average is zero. Hence
$$
\boxed{\mathbb E[X_{t'}X_t]=0\qquad(t'\ne t).}
$$
This argument covers overlapping driver sets at separations one and two, not just the disjoint case at larger separation. At equal times the value is $2$, so the complete autocovariance is $2\delta_{tt'}$. It establishes absence of second-order serial correlation, not independence of returns.
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