It is not appropriate to work in the space $H^1_0(U)$ since nonzero boundary conditions are being considered. You will have to assume some regularity of the boundary of $U$.
One version of Green's theorem (see e.g. the appendix in Evans) is that $$- \int_U (\Delta u) v \, dx = \int_U Du \cdot Dv \, dx -\int_{\partial U} \frac{\partial u}{\partial \nu} v \, dS.$$
A weak solution to the problem at hand can be proposed by setting $-\Delta u = f$ in $U$ and $\dfrac{\partial u}{\partial \nu} = -u$ on $\partial U$ so that $$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} uv \, dS \quad \forall v \in H^1(U)$$
or a bit more precisely
$$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS \quad \forall v \in H^1(U)$$
where $\tr : H^1(U) \to L^2(\partial U)$ is the trace operator.
An appropriate bilinear form is thus given by $$B[u,v] = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS, \quad u,v \in H^1(U).$$
$B$ is clearly bounded. As far as coercivity goes, it may be helpful to use the Rellich-Kondrachov theorem. I can follow up with a hint if you like.
It remains to show that there is a constant $\alpha > 0$ with the property that $\|u\|_{H^1}^2 \le \alpha B[u,u]$ for all $u \in H^1(U)$. This can be proven by contradition. Otherwise, for every $n \ge \mathbb N$ there would exist $u_n \in H^1(U)$ with the property that $\|u_n\|^2_{H^1} > n B[u_n,u_n]$. For each $n$ define $v_n = \dfrac{u_n}{\|u_n\|_{H^1}}$. Then $v_n \in H^1(U)$, $\|v_n\|_{H^1} = 1$, and $B[v_n,v_n] < \dfrac 1n$ and all $n$.
Here we can invoke Rellich-Kondrachov. Since the family $\{v_n\}$ is bounded in the $H^1$ norm, there is a subsequence $\{v_{n_k}\}$ that converges to a limit $v \in L^2(U)$. However, since $\|Dv_{n_k}\|_{L^2}^2 < \dfrac{1}{n_k}$ it is also true that $Dv_{n_k} \to 0$ in $L^2$. Thus for any $\phi \in C_0^\infty(U)$ you have $$\int_U v D \phi \, dx = \lim_{k \to \infty} \int_U v_{n_k} D \phi \, dx = - \lim_{k \to \infty} \int_U D v_{n_k} \phi \, dx = 0.$$ This means $v \in H^1(U)$ and $D v = 0$, from which you can conclude $v_{n_k} \to v$ in $H^1(U)$. Since $\|v_{n_k}\|_{H^1} = 1$ for all $k$ it follows that $\|v\|_{H^1} = 1$ as well.
Next, since $\|\tr v_{n_k}\|_{L^2(\partial U)}^2 < \dfrac{1}{n_k}$ and the trace operator is bounded there is a constant $C$ for which
$$ \|\tr v\|_{L^2(\partial \Omega)} \le \|\tr v - \tr v_{n_k}\|_{L^2(\partial \Omega)} + \|\tr v_{n_k}\|_{L^2(\partial \Omega)} < \frac{1}{n_k} + C \|v - v_{n_k}\|_{H^1(U)}.$$
Let $k \to \infty$ to find that $\tr v = 0$ in $L^2(\partial U)$.
Can you prove that if $v \in H^1(U)$, $Dv = 0$, and $\tr v = 0$, then $v = 0$? Once you have established that fact you arrive at a contradiction, since $v$ also satisfies $\|v\|_{H^1} = 1$. It follows that $B$ is in fact coercive.
