= Solution
The <Rellich-Kondrashov compactness theorem for H01> states that if $\Omega\subset\mathbb R^d$ is bounded and open, then the inclusion
$$
H_0^1(\Omega)\hookrightarrow L^2(\Omega)
$$
is compact: every bounded sequence in $H_0^1(\Omega)$ has a subsequence converging strongly in $L^2(\Omega)$.
Let $(u_n)$ be bounded in $H_0^1(\Omega)$. Since this is a Hilbert space, Banach--Alaoglu and reflexivity give a subsequence, still denoted $(u_n)$, and $u\in H_0^1(\Omega)$ such that $u_n\rightharpoonup u$ weakly in $H_0^1(\Omega)$ and hence in $L^2(\Omega)$. Extend all these functions by zero to $\mathbb R^d$. The zero extensions belong to $H^1(\mathbb R^d)$ and remain uniformly bounded there.
For each $\xi\in\mathbb R^d$,
$$
\widehat u_n(\xi)=\int_\Omega u_n(x)e^{-ix\cdot\xi}\,dx
\longrightarrow
\int_\Omega u(x)e^{-ix\cdot\xi}\,dx=\widehat u(\xi),
$$
because $e^{ix\cdot\xi}\mathbf1_\Omega\in L^2(\Omega)$. On each ball $|\xi|\leq R$, Cauchy--Schwarz gives a uniform bound on $|\widehat u_n-\widehat u|$, so dominated convergence yields
$$
\int_{|\xi|\leq R}|\widehat u_n-\widehat u|^2\,d\xi\longrightarrow0.
$$
The gradient bound controls high frequencies. By Plancherel,
$$
\int_{|\xi|>R}|\widehat u_n-\widehat u|^2\,d\xi
\leq \frac1{R^2}
\int_{\mathbb R^d}|\xi|^2|\widehat u_n-\widehat u|^2\,d\xi
\leq\frac C{R^2},
$$
uniformly in $n$. First choose $R$ large and then $n$ large. The low- and high-frequency estimates show $\widehat u_n\to\widehat u$ in $L^2(\mathbb R^d)$, and a final application of Plancherel gives $u_n\to u$ strongly in $L^2(\Omega)$.
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