= Solution
The <Nadaraya–Watson estimator> is
$$
\boxed{\widehat m_n(h,x)=\frac{\sum_{i=1}^nY_iK((x-X_i)/h)}{\sum_{i=1}^nK((x-X_i)/h)}.}
$$
For the <unit-width box kernel>, it averages the responses with design points in $[x-h/2,x+h/2]$. Define it to be zero if that window contains no observations; the supplied small-denominator bound must refer to a defined estimator.
Write $g=f^X$, $D_n=(nh)^{-1}\sum_i\mathbf1_{\{|X_i-x|\le h/2\}}$ and $U_n=(nh)^{-1}\sum_iY_i\mathbf1_{\{|X_i-x|\le h/2\}}$. Set $G_n=U_n-m(x)D_n$. On $D_n>\delta$,
$$
|\widehat m_n-m(x)|=|G_n|/D_n\le|G_n|/\delta.
$$
This is the <absolute error bound for a ratio estimator with a controlled denominator>; it does not require the numerator and denominator to be independent.
The <conditional expectation> definition of the <regression function> gives
$$
\mathbb EG_n=\frac1h\int_{-h/2}^{h/2}(m(x+u)-m(x))g(x+u)\,du.
$$
At this interior point, the <Taylor theorem> expansions are $m(x+u)-m(x)=m'(x)u+\tfrac12m''(x)u^2+o(u^2)$ and $g(x+u)=g(x)+g'(x)u+o(|u|)$. Multiplying and using the symmetric interval removes the odd linear term. More precisely,
$$
\mathbb EG_n=\frac{h^2}{12}\left(m'(x)g'(x)+\tfrac12m''(x)g(x)\right)+o(h^2)=O(h^2).
$$
This cancellation underlies the <interior absolute-error rate of the Nadaraya-Watson estimator>.
For the variance, <independence> between observations gives
$$
\operatorname{Var}(G_n)\le\frac1{nh^2}\mathbb E[(Y-m(x))^2\mathbf1_{\{|X-x|\le h/2\}}].
$$
Conditional on $X=t$, the second moment is $V(t)+(m(t)-m(x))^2$. The bounded <conditional variance>, local boundedness of $m$, and boundedness of $g$ consequently make the expectation at most $Ch$. Thus $\operatorname{Var}(G_n)=O((nh)^{-1})$. By the <Cauchy-Schwarz inequality>,
$$
\mathbb E[|\widehat m_n-m(x)|\mathbf1_{\{D_n>\delta\}}]
\le\delta^{-1}\left(|\mathbb EG_n|+\sqrt{\operatorname{Var}(G_n)}\right)
=O(h^2+(nh)^{-1/2}).
$$
The auxiliary result supplied in the question makes the expected error on $D_n\le\delta$ equal to $o(n^{-2/5})$. Combining both events, with $h_n\asymp n^{-1/5}$, proves
$$
\boxed{\mathbb E|\widehat m_n(h_n,x)-m(x)|=O(n^{-2/5}).}
$$
Both the squared-bandwidth bias and the inverse-square-root window-count error have this order. The supplied separate consistency result for the design-density estimator is not needed once the stronger small-denominator expectation bound is used.
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