= Solution
For smooth $u$, the fundamental theorem along the segment in direction $e_k$ gives, for either sign of $h$,
$$
\Delta_k^h u(x)=\int_0^1D_k u(x+the_k)\,dt.
$$
The entire segment lies in $\Omega$ under the distance hypothesis. The <Cauchy-Schwarz inequality> on $[0,1]$, followed by integration and translation, yields
$$
\int_{\Omega'}|\Delta_k^h u|^2\leq\int_0^1\int_{\Omega'}|D_k u(x+the_k)|^2\,dx\,dt
\leq\int_\Omega|D_k u|^2\leq\int_\Omega|Du|^2.
$$
For general $u\in W^{1,2}(\Omega)$, extend $u$ and its weak <gradient> separately by zero as $L^2$ functions on $\mathbb R^n$, and mollify $u$ with radius $\varepsilon$. At points farther than $\varepsilon$ from the boundary the mollified <gradient> equals the <convolution> of the zero-extended <gradient>, because the <convolution> kernel is supported wholly inside $\Omega$. Choose $\varepsilon<\operatorname{dist}(\Omega',\partial\Omega)-|h|$. All the segments then lie in this region, and the preceding calculation, followed by the $L^2$ contraction bound for <convolution>, gives the same right side $\|D_k u\|_{L^2(\Omega)}$. The <mollifications> converge locally in $L^2$, as do their fixed-$h$ translated differences, so passage to the limit preserves the estimate. Hence
$$
\boxed{\Delta_k^h u\in L^2(\Omega'),\qquad\|\Delta_k^h u\|_{L^2(\Omega')}\leq\|Du\|_{L^2(\Omega)}.}
$$
The argument proves the sharper estimate with $D_k u$ on the right. It is the basic Sobolev <difference quotient> bound.
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