= Solution
We prove <finite propagation speed> using a <shrinking cone energy argument>. Fix $T>0$ and $x_*$, and suppose the <Cauchy data> vanish on $B(x_*,T)$. For $0\leq s<T$, define the local <wave energy>
$$
e=\frac12\bigl(\phi_t^2+|\nabla\phi|^2\bigr),\qquad
E(s)=\int_{B(x_*,T-s)}e(s,x)\,dx.
$$
The homogeneous <wave equation> gives the local <conservation law>
$$
\partial_t e=\nabla\cdot(\phi_t\nabla\phi).
$$
Differentiate the integral over the moving ball. Its boundary moves inward with speed one, so the <divergence theorem> gives
$$
E'(s)=\int_{\partial B(x_*,T-s)}\bigl(\phi_t\partial_n\phi-e\bigr)\,dS
=-\frac12\int_{\partial B(x_*,T-s)}
\left((\phi_t-\partial_n\phi)^2+|\nabla_{\mathrm{tan}}\phi|^2\right)\,dS\leq0.
$$
Here $n$ is the outward <unit normal>, $\partial_n$ the <normal derivative>, and $\nabla_{\mathrm{tan}}$ the component of the <gradient> tangent to the boundary. Since $E(0)=0$ and $e\geq0$, we have $E(s)=0$. Hence both $\phi_t$ and $\nabla\phi$ vanish inside the backward <light cone>. Integrating $\phi_t(s,x_*)=0$ from the zero initial displacement gives $\phi(s,x_*)=0$; <continuity> then gives $\phi(T,x_*)=0$.
Applying the same <energy estimate> to the difference of two solutions proves the <domain of dependence> assertion. If $\operatorname{dist}(x_*,K)>T$, the initial ball misses $K$, and the preceding argument proves the stated <support> bound. \b[No disturbance propagates faster than one.]
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