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.
Here is a solution more related to Evan's approach in the mentioned textbook. Let $u = u(x,t)$ be a solution of the PDE. Then using integration by parts and that $u_t = \Delta u$ we get
$$\begin{align}\frac{d}{dt} \left(\frac{1}{2} \|u\|_{L^2(U)}^2\right) &= \int_U u_tu\ dx = \int_U u \Delta u\ dx\\&= -\int_U |Du|^2 dx \overset{(\ast)}\leq -
\lambda_1 \|u\|^2_{L^2(U)}\end{align}$$
where $(\ast)$ comes from Rayleigh's Formula
$$\lambda_1 = \underset{\substack {u \in H_0^1 (U)\\ u\neq 0}}\min \frac{B[u,u]}{\|u\|^2_{L^2(U)}} = \underset{\substack{u \in H_0^1 (U)\\ u\neq 0}} \min \frac{\int_U |Du|^2 dx}{\|u\|^2_{L^2(U)}} $$
Now let $\eta (s) = \|u(\cdot,s)\|^2_{L^2(U)}$. Then
$$\frac{d}{ds} \left(\eta(s) e^{2\lambda_1 s}\right) = e^{2\lambda_1 s} (\eta'(s) +2\lambda_1 \eta(s)) \leq 0$$
Integrating from $0$ to $t$ w.r.t. $s$ we obtain
$$\eta(t)e^{2\lambda_1 t} \leq \eta (0)$$
Since $\eta (0) = \|u (\cdot, 0)\|^2_{L^2(U)} = \|g\|^2_{L^2(U)}$ the result follows.
Best Answer
I think you have the right notion, and in Exercise 5, things are 'nicer' for us in that there is no time variable to deal with.
In the following, I am assuming that the given $f \in L^2(U)$.
\begin{align} -\Delta w &= g \quad \text{in } U\\ w &= 0 \quad \text{on } \partial U. \end{align}
By appealing to existence and regularity results (see Chapter 6 in Evans), we know that there exists a unique weak solution $w \in H^1_0(U)\cap H^2(U)$ to the above linear PDE.
Define an operator $A: H^1_0(U) \to H^1_0(U)$ by \begin{equation} A[u] = w \end{equation} Let $u,\tilde{u} \in H^1_0(U)$, with $A[u] = w, A[\tilde{u}] = \tilde{w}$. To show: $A$ is a strict contraction. (This is a good place to try yourself first without reading further). Then
\begin{align} \left\|A[u] - A[\tilde{u}]\right\|_{H^1_0(U)} &= \|w-\tilde{w}\|_{H^1_0(U)}\\ &\leq C\|D(w - \tilde{w})\|_{L^2(U)}\quad(*).\end{align} In the above, $(*)$ follows by a Sobolev estimate. Now \begin{align} \|D(w - \tilde{w})\|_{L^2(U)}^2 &= \int_{U} D(w - \tilde{w})\cdot D(w - \tilde{w}) dx\\ &= -\int_{U}[\Delta(w - \tilde{w})] (w - \tilde{w}) dx\\ &= \int_{U} (b(D\tilde{u}) - b(Du))(w - \tilde{w}) dx\\ &\leq \|b(D\tilde{u}) - b(Du)\|_{L^2(U)}\|w - \tilde{w}\|_{L^2(U)}\quad (**)\\ &\leq C\|b(D\tilde{u}) - b(Du)\|_{L^2(U)}\|D(w - \tilde{w})\|_{L^2(U)} \quad (***) \end{align} In $(**)$, I have used the fact that $b$ is Lipschitz so $b(Du), b(D\tilde{u}) \in L^2(U)$. In $(***)$, I have used a Sobolev estimate. Our conclusion, having reached $(***)$, is that \begin{equation} \|D(w - \tilde{w})\|_{L^2(U)} \leq C\|b(D\tilde{u}) - b(Du)\|_{L^2(U)}. \end{equation} Hence returning to $(*)$, we see that \begin{align} \left\|A[u] - A[\tilde{u}]\right\|_{H^1_0(U)} &\leq C\|b(D\tilde{u}) - b(Du)\|_{L^2(U)}\\ &\leq C\text{Lip}(b)\|D\tilde{u} - Du\|_{L^2(U)}\\ &\leq C\text{Lip}(b)\|u - \tilde{u}\|_{H^1_0(U)}. \end{align} Provided $\text{Lip}(b)$ is small enough, $A$ is a strict contraction (here the constant $C$ is from the Sobolev estimates, so only depends on things like the dimension $n$, $U$, etc.).
Thus by the Banach fixed point theorem, we deduce the existence of a unique fixed point $u_* \in H^1_0(U)$, such that $A[u_*] = u_*$. The regularity theory for the PDE in 1. gives $u_* \in H^2(U)\cap H^1_0(U)$.