= Solution
Let $A_n=\sum_{i=1}^{m_n}f(X_i)$ and $T_n=\sum_{i=m_n+1}^n f(X_i)$. Since $|A_n|\leq Bm_n$, $\operatorname{Var}(A_n)\leq B^2m_n^2$. The <Cauchy-Schwarz inequality> applied to the centered sums gives
$$
\sqrt{\operatorname{Var}(A_n+T_n)}
\leq\sqrt{\operatorname{Var}(A_n)}+\sqrt{\operatorname{Var}(T_n)}.
$$
Divide by $\sqrt n$. The first term is at most $Bm_n/\sqrt n\to0$, while part (b) and $(n-m_n)/n\to1$ give a limsup of at most $\sqrt{\gamma\operatorname{Var}_\mu(f)}$ for the second. Squaring proves
$$
\boxed{\limsup_{n\to\infty}\frac1n\operatorname{Var}\!\left(\sum_{i=1}^n f(X_i)\right)\leq\gamma\operatorname{Var}(f(Y)).}
$$
This argument also covers $\operatorname{Var}_\mu(f)=0$; it does not require independence between the discarded prefix and the retained path. The same chain-dependent $\gamma$ as in part (b) suffices.
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