= Solution
The boundedness in the <Sobolev space> $H^1(U)$ and the <weak sequential compactness of bounded sequences in a reflexive Banach space> give a subsequence converging weakly to some $u\in H^1(U)$. For each integer $m\geq2$, the <Rellich-Kondrachov compactness theorem> makes $H^1(U)\hookrightarrow L^m(U)$ compact because the dimension is two. Repeated extraction followed by the <diagonal argument> gives one subsequence $u_{n_k}$ converging strongly to $u$ in every $L^m(U)$ with integral $m\geq2$.
For any finite real $p\geq1$, choose an integer $m\geq\max\{2,p\}$. Since $U$ has finite measure, the <Lp inclusion on a finite measure space> gives
$$
\|u_{n_k}-u\|_{L^p(U)}
\leq |U|^{1/p-1/m}\|u_{n_k}-u\|_{L^m(U)}\longrightarrow0.
$$
The same subsequence therefore works for every finite $p$.
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