= Solution
Work with real-valued smooth functions on the bounded region enclosed by $S$, with the <oriented surface element> pointing outward. Set $\psi=\phi-\phi_1$. It is a <harmonic function> in $V$ and has zero <Dirichlet boundary conditions> on $S$. Applying <Green's first identity> to $\psi$ with itself gives
$$
\int_V|\nabla\psi|^2\,dV
=\int_S\psi\,\partial_n\psi\,dS-\int_V\psi\Delta\psi\,dV=0.
$$
Both terms on the right vanish by the boundary condition and harmonicity. The integrand is nonnegative and continuous, so $\nabla\psi=0$ throughout $V$. Consequently $\psi$ is constant on each <connected component>; every bounded component meets the prescribed boundary, where that constant is zero. Therefore
$$
\boxed{\phi_1=\phi\text{ throughout }V.}
$$
This is the energy proof of <Uniqueness of the Dirichlet problem>. Connectedness of $V$ is not essential if the boundary values are prescribed on every component. Boundedness, or suitable decay and integrability at infinity, is needed for the boundary/energy argument; the finite enclosed-volume interpretation is used here.
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