= Solution
The nonexistence conclusion concerns nonconstant fields: a constant <scalar-field vacuum> with $U=0$ is a zero-energy <critical point of an energy functional> in every dimension. Assume the potential is smooth, so the classical <Euler-Lagrange equation> is well defined. Write the two nonnegative <energy> terms as
$$
T=\frac12\int|\nabla\phi|^2\,d^Dx,\qquad
V=\int U(\phi)\,d^Dx.
$$
For the admissible <Derrick scaling> $\phi_\lambda(x)=\phi(\lambda x)$,
$$
E(\phi_\lambda)=\lambda^{2-D}T+\lambda^{-D}V.
$$
Stationarity at $\lambda=1$ gives the <Derrick virial identity>
$$
\boxed{(2-D)T-DV=0.}
$$
For $D>2$, both terms have nonpositive sign, forcing $T=V=0$ and hence a constant vacuum. For $D=2$, scaling alone only proves $V=0$; the gradient term is scale invariant, so it must not simply be declared zero.
The <two-dimensional flat-target Derrick obstruction> supplies that remaining step. Since $U\geq0$ and its integral is zero, continuity gives $U(\phi(x))=0$ everywhere; smooth nonnegative $U$ has $U'(\phi(x))=0$ at those values. The field equation $\Delta\phi=U'(\phi)$ therefore becomes $\Delta\phi=0$. Each $\partial_i\phi$ is an entire <harmonic function> in $L^2(\mathbb R^2)$. Its <mean value property> and the <Cauchy-Schwarz inequality> give, for every radius $R$,
$$
|\partial_i\phi(x_0)|
\leq\frac{\|\partial_i\phi\|_{L^2(\mathbb R^2)}}{\sqrt{\pi R^2}}.
$$
Let $R\to\infty$ to get $\partial_i\phi=0$. Thus $D=2$ also has only constant <stationary field configurations> of finite <energy>. The unconstrained real scalar target is essential in this argument; the spherical target in Question 2 has different field equations.
In $D=1$ the field equation is $\phi''=U'(\phi)$. Multiplication by $\phi'$ gives the <first integral>
$$
\frac12(\phi')^2-U(\phi)=K.
$$
Finite <energy> makes $\frac12(\phi')^2+U(\phi)$ an <integrable> nonnegative function, so along some sequence $x_n\to+\infty$ it tends to zero. Evaluating the constant <first integral> on that sequence gives $K=0$. Choose
$$
\boxed{W(u)=\int_{u_*}^{u}\sqrt{2U(s)}\,ds,\qquad U(u)=\frac12(W'(u))^2.}
$$
Consequently $(\phi')^2=2U(\phi)$ and
$$
\boxed{\frac{d\phi}{dx}=\pm\frac{dW}{d\phi}.}
$$
For a nonconstant entire solution the sign is fixed. Indeed a finite-$x$ zero of $\phi'$ would have $U(\phi)=U'(\phi)=0$; uniqueness of the smooth second-order equation with initial data $(\phi,\phi')=(v,0)$ would make the solution the constant vacuum. Thus $\phi'$ never vanishes in a nonconstant solution. Conversely a smooth solution of the displayed first-order equation satisfies $\phi''=U'(\phi)$ wherever it is nonconstant.
Under the usual soliton boundary conditions $\phi(x)\to v_\pm$ as $x\to\pm\infty$, with $v_\pm$ in the <vacuum manifold> $\{U=0\}$, the two endpoint components label the <topological sector>. For isolated <scalar-field vacua> this is an ordered pair of vacuum labels, not necessarily one universally normalized integer. More explicitly, for any smooth $F$, the <topological current>
$$
j_F^\mu=\epsilon^{\mu\nu}\partial_\nu F(\phi),\qquad
\partial_\mu j_F^\mu=0,\qquad
Q_F=\int j_F^0\,dx=F(v_+)-F(v_-)
$$
is identically conserved, taking $\epsilon^{01}=1$ and boundary conditions fixed during evolution. The choices $F(u)=u$ and $F(u)=W(u)$ give the field-difference charge and the kink <energy> charge. The <square completion for a one-dimensional kink> gives
$$
E=\frac12\int(\phi'\mp W'(\phi))^2\,dx
\pm[W(v_+)-W(v_-)],
\qquad
\boxed{E\geq|W(v_+)-W(v_-)|.}
$$
The first-order critical solution saturates this <Bogomolny bound>. A nonzero difference between endpoint vacuum components prevents continuous deformation to a vacuum while those boundary conditions remain fixed. Finite <energy> alone need not give finite endpoint values for arbitrary potentials, so these topological labels use the stated vacuum boundary conditions.
To evade the scalar-only obstruction, introduce a <gauge field> and a charged, generally multicomponent <Higgs field>. For a Yang-Mills-Higgs <energy> write $E=T_H+V+E_B$, where $T_H$ contains $|D_i\Phi|^2$ and $E_B$ contains $|F_{ij}|^2$. Rescale $\Phi_\lambda(x)=\Phi(\lambda x)$ and $A_{i,\lambda}(x)=\lambda A_i(\lambda x)$. Then $D_i\Phi$ scales by $\lambda$, while $F_{ij}$ scales by $\lambda^2$, so
$$
E_\lambda=\lambda^{2-D}T_H+\lambda^{-D}V+\lambda^{4-D}E_B,
\qquad
\boxed{(2-D)T_H-DV+(4-D)E_B=0.}
$$
The magnetic term supplies the missing opposing scaling power. In $D=2$ the relation is $E_B=V$, permitting <Abelian Higgs vortices>; in $D=3$ it is $E_B=T_H+3V$, permitting <'t Hooft-Polyakov monopoles>. In the monopole Bogomolny limit $V=0$, $B_i=\pm D_i\Phi$ achieves $E_B=T_H$. Also, asymptotic ordinary derivatives need not vanish when a Higgs phase winds: the relevant vanishing quantity is the <gauge covariant derivative>. This allows nontrivial boundary winding with finite <energy>. Merely appending a gauge potential to a neutral one-component real field would not by itself furnish that Higgs mechanism.
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