Thanks for the hint in the comment. Instead of using Lax-Milgram, we should invoke Fredholm alternative(Theorem 4 of Chapter 6.2.3 in Evans). Consider the equation $$\begin{cases}
Lu = 0 \ \ \ \text{ in $U$} \\
u = 0 \ \ \ \ \ \ \text{ on $\partial U$}.
\end{cases}$$
Suppose $u$ is a weak solution to $Lu= 0$. Then note that by regularity theorem, $u \in C^\infty(\overline{U})$. By the strong maximum (and minimum) principle, $u \equiv 0$. Therefore, $Lu=0$ has only the trivial solution, which, by Fredholm alternative, says that for each $f \in X = H^m(U) \cap H_0^1(U)$, we have a unique solution $u \in H^1_0(U)$. Invoking again regularity theorem, $u \in H^{m+2}(U) \cap H^1_0(U) \subset X$.
The key fact that he's using without clearly stating it is that a uniformly continuous function on a dense subset of a metric space uniquely extends to a uniformly continuous function on the full metric space. If Evans assumed that $u$ were uniformly continuous on all of $U$, then the density of $U$ in $\bar{U}$ would give us the extension of $u$ to $\partial U$ immediately. However, he only assumes the local uniform continuity condition, so we need to do a bit more work. The idea is that if we fix a point $p \in \partial U$ and consider the ball $B(p,r)$, then $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$, and our function $u$ is uniformly continuous in the former set by assumption. So, we can extend it uniquely to a continuous function on $\bar{U} \cap B(p,r)$, which in particular allows us to uniquely assign values to $u$ on $\partial U \cap B(p,r)$. This works for all $p \in \partial U$, and so we can uniquely extend $u$ to $\partial U$ using this trickery, and the resulting extension is continuous.
On the other hand, you can ask about the space
$$
\dot{C}(\bar{U}) = \{u : \bar{U} \to \mathbb{R} \;\vert\; u \text{ is continuous on } \bar{U}\}
$$
(I've added the dot to distinguish this from his space). We've just seen that any $u \in C(\bar{U})$ (his space) uniquely extends to define an element of $\dot{C}(\bar{U})$. On the other hand, any $u \in \dot{C}(\bar{U})$ restricted to $U$ defines an element of $C(\bar{U})$ since $u$ will be uniformly continuous on compact subsets of $\bar{U}$, and any bounded subset of $U$ will be contained in such a compact subset.
Ultimately, what we've done is built a linear extension isomorphism (you can check easily that it's linear) $E : C(\bar{U}) \to \dot{C}(\bar{U})$, and a linear restriction isomorphism $R: \dot{C}(\bar{U}) \to C(\bar{U})$ such that $ER = I$ and $RE = I$. In other words, these spaces are naturally isomorphic via these extension and restriction processes. So, what's the point? In some sense the space $C(\bar{U})$ is preferable because it only requires the functions involved to be defined on $U$. This means that statements like $W^{k,p}(U) \hookrightarrow C(\bar{U})$ (for appropriate Sobolev parameters $k,p$) are slightly more natural because the functions in the Sobolev space are only defined on $U$ a priori. If we were to write $W^{k,p}(U) \hookrightarrow \dot{C}(\bar{U})$, then there would be a slight abuse of notation in that the functions in the first space are only defined a priori in the smaller space $U$. This isn't a big deal, as we've seen above with $E$, and many texts simply gloss over this issue and take the statement $W^{k,p}(U) \hookrightarrow \dot{C}(\bar{U})$ to mean
$$
W^{k,p}(U) \hookrightarrow C(\bar{U}) \xrightarrow{E} \dot{C}(\bar{U}).
$$
EDIT: Here is a proof of the claim that $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$ in any metric space. Let $\varepsilon >0$ and pick a point $y \in \bar{U} \cap B(p,r)$. We know that $\bar{U} = U \cup \partial U$. If $y \in U \cap B(p,r)$, then we simply pick $x = y \in U \cap B(p,r)$ and note that $d(x,y) = 0 < \varepsilon$. Suppose, then, that $y \in \partial U \cap B(p,r)$. By definition of the boundary, we can pick $x \in U \cap B(y,s)$ for $s = \min\{\varepsilon, r-d(y,p) \}$. Then, certainly, $d(x,y) < s \le \varepsilon$, but also $d(x,p) \le d(x,y) + d(y,p) < s + d(y,p) \le r$, so $x \in U \cap B(p,r)$. We have now shown that for every $y \in \bar{U} \cap B(p,r)$ and $\varepsilon >0$ there exists $x \in U \cap B(p,r)$ such that $d(x,y) < \varepsilon$, so $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$.
Best Answer
Hint: You can compute for all $\phi \in C_c^\infty(0,T)$ $$\begin{aligned} ((Eu)',\phi)_{L^2(0,T)}=-(Eu,\phi')_{L^2(0,T)}&=-\left(Eu,\lim_{h \to 0} \frac{\phi(t+h)-\phi(t)}{h}\right)_{L^2(0,T)} \\ &=\left(\lim_{h \to 0} \frac{Eu(t)-Eu(t-h)}{h},\phi\right)_{L^2(0,T)} \end{aligned}$$