Any Hilbert space is of course self dual.
But this is only true if you use the inner product as the pairing.
Note that $g\in H^{-1}$ acts on $f\in H^1_0$ (formally) by $g(f)=\int_U gf$, where no derivatives appear.
Often the dual space of a function space is considered to act by "a simple integral", which may not be the inner product (if we are not in $L^2$).
Let me try to make this a bit more concrete.
Let us write $\langle f,g\rangle=\int_Ufg$ and $(f,g)=\langle f,g\rangle+\langle \nabla f,\nabla g\rangle$.
The second one is the inner product on $H^1_0$.
Then $H^1_0\ni f\mapsto(f,\cdot)\in (H^1_0)'$ is surjective (as always in a Hilbert space), but $H^1_0\ni f\mapsto\langle f,\cdot\rangle\in (H^1_0)'$ is not.
We start with a theorem:
Theorem. Let $H$ be a Hilbert space and $T$ a compact self adjoint operator. Then there exist a Hilbert basis composed of eigenvectors of $T$.
Proof. See Brezis chapter 6, in particular, theorem 6.11.
Remark: Let $(\cdot,\cdot)$ denote inner product. In the conditions of the theorem, if $(Tu,u)> 0$ for all $u\neq 0$ then, all eigenvalues are positive.
Now, as $-\Delta$ is the model of an elliptic operator, let's assume that $L=-\Delta$. Consider the Dirichlet problem
$$\tag{1}
\left\{ \begin{array}{ccc}
-\Delta u=f &\mbox{ in $U$,} \\
u=0 &\mbox{on $\partial U$,}
\end{array} \right.
$$
where $f\in L^2(U)$. For each $f\in L^2(U)$, problem $(1)$ has a unique solution in $H_0^1(U)$ in the sense that $$\tag{2}\int_U\nabla u\nabla v=\int_U fv,\ \forall \ v\in H_0^1(U).$$
This can be proved for example, by minimising the functional $I:H_0^1(U)\to \mathbb{R}$, defined by $$I(u)=\frac{1}{2}\int_U |\nabla u|^2 .$$
Let $S:L^2(U)\to H_0^1(U)$ be the solution operator, i.e. for each $f\in L^2(U)$, $Sf$ is the unique solution of $(1)$.
It is straightforward to see that $S$ is linear. Once $H_0^1(U)$ is compactly embedded in $L^2(U)$ (Rellich-Kondrachov theorem), we have that $S:L^2(U)\to L^2(U)$ is a compact operator. Moreover, $S$ is symmetric (in $L^2(U)$) because $$(Sf,g)=\int_U (Sf)g=\int_U \nabla (Sf)\nabla (Sg)=\int_U fSg=(f,Sg),\ \forall\ f,g\in L^2(U).$$
In the above, we have used $(2)$. To conlude, note also that $(2)$ implies that $S$ is positive: $$(Sf,f)=\int_U (Sf)f=\int_U |\nabla (Sf)|^2>0,\ \forall\ f\in L^2(U),\ f\neq 0.$$
We apply the theorem, followed by the remark, to find a sequence $(\lambda_n,\phi_n)\in (0,\infty)\times L^2(U)$ satisfying $$S\phi_n=\lambda_n\phi_n,\ \forall n,\tag{3}$$
$\phi_n$ is orthonormal, $\lambda_n\to 0$. Once $(3)$ is satisfied, we conclude that each $\phi_n$ is in $H_0^1(U)$. Note that $-\Delta:H_0^1(U)\to H^{-1}(U)$ is a linear operator defined by $$\langle -\Delta u,v\rangle =\int_U \nabla u\nabla v,$$
where $H^{-1}(U)$ is the dual of $H_0^1(U)$, $\langle\cdot,\cdot\rangle$ denotes duality. So, as you can see from the definition of $-\Delta$, $(3)$ implies that
$$
\left\{ \begin{array}{ccc}
-\Delta \phi_n=\frac{1}{\lambda_n}\phi_n &\mbox{ in $U$,} \\
\phi_n=0 &\mbox{on $\partial U$,}
\end{array} \right.
$$
which conclude the proof of Evans theorem.
Best Answer
Consider the norm $$\| u\| := \|\nabla u \|_{L^2(U)}=\bigg ( \int_U \vert \nabla u \vert^2 \, dx \bigg )^{1/2} $$ on $H^1_0(U)$ (which is equivalent to the usual norm on $H^1_0(U)$ by the Poincaré inequality) and let $$ \| u \|':= (B[u,u])^{1/2}.$$ Then $\|\cdot \|$ and $\|\cdot \|'$ are equivalent norms on $H^1_0(U)$.
Indeed, Evans assumes $a^{ij}$ are bounded so for all $u,v\in H^1_0(U)$, $$B[u,u] = \sum_{i,j}\int_U a^{ij}u_{x_i}u_{x_j}\,dx\leqslant C\sum_{i,j}\int_U \vert u_{x_i}\vert \cdot \vert u_{x_j}\vert\,dx \leqslant C \| \nabla u \|_{L^2(U)}^2 $$ Moreover, by ellipticity, $$\sum_{i,j}\int_U a^{ij}u_{x_i}u_{x_j}\,dx \geqslant \theta \| \nabla u \|^2_{L^2(U)} . $$
This immediately implies that $(H^1_0(U), \| \cdot \|')$ is a Banach space with the same topology as $H^1_0(U)$ (with the usual norm). But then define $(u,v)':= B[u,v]$. One can verify that $(H^1_0(U), ( \cdot,\cdot ) ')$ is an inner product space and from the definition of $\| \cdot \|'$, the inner product $( \cdot,\cdot ) '$ induces the norm $\| \cdot \|'$. Then $(H^1_0(U), ( \cdot,\cdot ) ')$ is a Hilbert space since $\| \cdot\|'$ is complete on $H^1_0(U)$.