Suppose $u$ is smooth so that $u_{x_i t} = u_{tx_i}$ where $x_i$ is any spacial variable, and the energy is bounded, then we can put the differentiation inside the integral sign:
$$
\frac{dE}{dt} = \iiint_D \frac{d}{dt}(u_{t}^2 + c^2|\nabla u|^2) dx
$$
use the product rule:
$$
\frac{dE}{dt} = \iiint_D \big(2u_t u_{tt} + 2c^2 \nabla u\cdot (\nabla u)_t\big)dx
$$
Now $u_{tt} = c^2\Delta u$, hence
$$
\frac{dE}{dt} = \iiint_D \big(2u_t c^2\Delta u + 2c^2 \nabla u\cdot (\nabla u)_t\big)dx
$$
Use Green's identities (integration by parts using divergence theorem):
$$\iiint_D \psi \Delta \varphi \,dx = -\iiint \nabla \varphi \cdot \nabla \psi\, dx +\iint_{\partial D} \psi ( \nabla \varphi \cdot \boldsymbol{n} )\, dS $$
where $\psi = u_t$, and $\varphi = u$
$$
\frac{dE}{dt} = \iiint_D \big(-2c^2\nabla u\cdot \nabla(u_t) +2 c^2 \nabla u\cdot (\nabla u)_t\big)dx + \iint_{\partial D} u_t ( \nabla u \cdot \boldsymbol{n} )\, dS
$$
For the boundary part, for $u = 0$ on boundary for all time, hence
$$
u_t ( \nabla u \cdot \boldsymbol{n} ) = u_t ( \nabla u \cdot \boldsymbol{n} ) +
u ( \nabla u \cdot \boldsymbol{n} )_t = \big(u (\nabla u \cdot \boldsymbol{n} )\big)_t = 0
$$
for $u (\nabla u \cdot \boldsymbol{n} )$ is always zero on boundary regardless of time. Therefore:
$$
\frac{dE}{dt} = \iiint_D \big(-2c^2\nabla u\cdot \nabla(u_t) +2 c^2 \nabla u\cdot (\nabla u)_t\big)dx
$$
Now that $u$ is smooth thus $\nabla(u_t) = (\nabla u)_t$, we have
$$
\frac{dE}{dt} = 0
$$
and the conservation of energy is proved.
