As dls mentioned you're lacking an integral on the RHS. With this notice that
$$
\sum_{i,j=1}^n \partial_i(a_{ij} \partial_j u) =\text{div}(a \nabla u),
$$
so that, integrating by parts and using the zero boundary conditions, we get that the RHS is equal to
$$
-\int_{\Omega} \langle \nabla u, a \nabla u \rangle dx \leq 0.
$$
Edit: Notice that you want to prove that if $u(x,0)=0$ for every $x\in \Omega$ and $u(y,t)=0$ for every $y\in \partial \Omega$ and $t>0$ then $u\equiv 0$. So take such an $u$, then your argument gives
$$
\frac{1}{2}\frac{d}{dt} \int_{\Omega} |u|^2 dx = \int_{\Omega} u \text{div}(a\nabla u) dx.
$$
Call $v=a\nabla u$. The divergence theorem, applied to the vector field $uv$, gives
$$
\int_{\Omega}u \text{div}(v)dx= -\int_{\Omega}\langle \nabla u, v \rangle dx + \int_{\partial \Omega}\langle uv,\nu \rangle d \sigma.
$$
Since we have that $u(y,t)=0$ for every $y\in \partial \Omega$ the boundary integral is zero for every $t$. Therefore we have
$$
\frac{1}{2}\frac{d}{dt}\int_{\Omega} |u|^2 dx =- \int_{\Omega} \langle \nabla u,a\nabla u\rangle dx \leq 0.
$$
I think you need a $+$ in your energy rather than a minus, to account for an integration by parts.
As you said, suppose $U_1$ and $U_2$ are two solutions and take the difference $w = U_1 - U_2$. Then $w$ solves a very similar problem
$$w_{tt} + w_t - c^2w_{xx} = 0$$
$$w(a,t) = w(b,t)=0$$
$$w(x,0) = 0$$
$$w_t(x,0) = 0$$
Notice that as a consequence of the boundary conditions $w_t(a,t) = 0$ and $w_t(b,t) = 0$, since $w(a, \cdot)$ (resp. $b$) is always $0$. Notice also for my suggested energy $E(t) = \int_a^b (w_t)^2 + c^2(w_x)^2 \, dx$, if $E(t) = 0$ for all $t$ then $w_t =0$ for all $t$, thus $w(x, \cdot)$ is constant and $w(x,0)=0$, therefore $w = 0$.
Differentiate $E$,
$$E'(t) = \frac{1}{2}\int_a^b \left( 2 w_t w_{tt} + 2 c^2 w_x w_{x,t} \right) \, dx$$
We would like to use the equation $w$ satisfies to make this $0$ or $\leq 0$. The obstacles are that the $x$-derivative terms are not quite right, and the absence of an "independent" $w_t$ term.
Integrate $\int w_x w_{x,t} \, dx$ by parts, using the fact that $w_t(a,t)=w_t(b,t)=0$ to dispose of the boundary terms leaving only $-\int w_{xx} w_t \, dx$. This places us at
$$E'(t) = \int_a^b w_{tt} w_t - c^2 w_{xx} w_t \, dx$$
Add and subtract $(w_t)^2$ inside the integral,
$$E'(t) = \int_a^b (w_{tt} w_t + w_t w_t - c^2 w_{xx} w_t) - (w_t)^2 \, dx$$
Using $w_{tt} + w_t - c^2 w_{xx} = 0$, this ends up at $E'(t) = \int_a^b - (w_t)^2 \, dx$, which is a negative quantity. Therefore $E$ is decreasing. Also Observe that $E(0) = 0$ and $E(t) \geq 0$ to conclude that $E(t)$ is constantly $0$.
Best Answer
The intuition is obvious: It's the heat equation. The first boundary condition says that the region $\Omega$ starts with a temperature of $0$. And the boundary of $\Omega$ is kept at a temperature of $0$ for all time. So the temperature must be $0$ for all time.
You almost have the argument. $\mathcal E$ is an integral of an everywhere non-negative function, so it's always non-negative. It starts at $0$ and is non-increasing and non-negative, so the only thing it can do is stay at $0$. Now $w$ is something whose square integral is $0$ and it's at least continuous, so $w = 0$.