OK, perhaps the notation in Ref. 1 is a bit confusing. Let us elaborate on Derrick's No-Go theorem:
Derrick's No-Go theorem: For the number of spatial dimensions $D>2$, the only time-independent finite-energy solutions are ground states.
In a nutshell, the idea of the proof is to derive a necessary condition by a simple 1-parameter scaling argument. (Contrary to what one might naively expect, it is not particularly useful to set up Euler-Lagrange (EL) equations via functional differentiation because we do not know much about the potential $U$.)
Since we are considering time-independent field configurations, we should find stationary$^1$ field configurations for the total potential energy
$$V[\phi]~=~V_1[\phi]+ V_2[\phi], \tag{1}$$
where
$$\begin{align} V_1[\phi] ~:=&~\frac{1}{2} \int \!\mathrm{d}^Dx~ g_{\alpha\beta}(\phi)\partial_i\phi^{\alpha}\partial^i\phi^{\beta} ~\geq~ 0, \cr V_2[\phi]~:=~& \int \!\mathrm{d}^Dx~ U(\phi)~\geq~ 0, \cr U(\phi) ~\geq~& 0,\end{align} \tag{2}$$
and where $g_{\alpha\beta}$ is a positive definite target space metric.
Now we want to check if some field configuration ${\bf x} \mapsto \phi_1({\bf x})$ is stationary. Define a 1-parameter family of field configurations
$$ \phi_{\lambda}({\bf x})~:=~\phi_1(\lambda{\bf x}), \qquad \lambda ~\in~ [0,\infty[.\tag{3} $$
Note that $\phi_{\lambda=1}=\phi_1$, so the notation (3) is consistent. We then calculate (by change of variables in the integrals) that
$$ V[\phi_{\lambda}]~=~\lambda^{2-D}V_1[\phi_1]+\lambda^{-D}V_2[\phi_1].\tag{4}$$
A necessary (but far from sufficient!) condition for $\phi_1$ to be stationary is to differentiate the function $\lambda \mapsto V[\phi_{\lambda}]$ wrt. the parameter $\lambda$ at the point $\lambda=1$,
$$ 0~\stackrel{?}{=}~ \left. \frac{d V[\phi_{\lambda}]}{d\lambda}\right|_{\lambda=1}~=~ (2-D) V_1[\phi_1] - D V_2[\phi_1]. \tag{5}$$
Varying along the 1-parameter family constitutes only one out of infinitely many possibilities to vary the field configuration $\phi_1$, but it is the only one we'll need!
We next assume$^2$ that $D>2$. Then eq. (5) is only possible if
$$ V_1[\phi_1]~=~0~=~V_2[\phi_1].\tag{6} $$
Eq. (6a) implies that $$ \partial \phi_1 \equiv 0,\tag{7} $$
or equivalently that
$$ \phi_1 \text{ is ${\bf x}$-independent}.\tag{8} $$
Eq. (6b) and (8) then imply that
$$ \phi_1 \text{ is a ground state}.\tag{9} $$
References:
S. Coleman, Aspects of Symmetry, 1985; p. 194.
R. Rajaraman, Solitons and Instantons: An Introduction to Solitons and Instantons in Quantum Field Theory, 1987; Section 3.2 & 3.3.
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$^1$ Stationary in the sense that they satisfy EL equations for $V$, i.e. the functional derivative vanishes,
$$0~\approx~\frac{\delta V[\phi]}{\delta \phi^{\gamma}}~=~ -g_{\gamma\beta}(\phi)\partial^2\phi^{\beta} - \Gamma_{\gamma,\alpha\beta}(\phi)\partial_i\phi^{\alpha}\partial^i\phi^{\beta} + \frac{\partial U(\phi)}{\partial \phi^{\gamma}}.\tag{10} $$
[Here the $\approx$ symbol means equality modulo EL equations.]
$^2$ The case $D=2$. Contrary to what Ref. 1 claims, the Derrick's No-Go theorem does not hold for $D=2$. Let us here consider just the $D=2$ case.
We claim that one can no longer conclude that $V_1[\phi_1]$ should be zero. Ref. 1 gives the following wrong proof (using our notation):
For $D = 2$, however, Eq. (5) only implies the vanishing of $V_2[\phi_1]$, and a small amount of further argument is required. If $V_2[\phi_1]$ vanishes it is stationary, since zero is its minimum value. Thus we may apply Hamilton's principle to $V_1[\phi_1]$ alone, from which it trivially follows that $V_1[\phi]$ also vanishes. Q.E.D.
The potential energy term $V_2[\phi_1]=0$ must still be zero, cf. eq. (5). In other words, the $\phi_1$-image
$$ {\rm Im}(\phi_1) ~\subseteq ~U^{-1}(\{0\}) \tag{11} $$
must lie in the zero locus $U^{-1}(\{0\})$ (or preimage of $\{0\}$) of the potential $U$, i.e. the set of minimum points for the potential $U$ in target space.
Lagrange multiplier. As an aside, if a constraint $\chi(\phi)\approx 0$ emulates the zero locus $U^{-1}(\{0\})=\chi^{-1}(\{0\})$, one may effectively replace the functional (1) with
$$\tilde{V}[\phi,\Lambda]~=~V_1[\phi]+ \int \!\mathrm{d}^Dx~\Lambda~ \chi(\phi) , \tag{12} $$
where $\Lambda=\Lambda({\bf x})$ is a Lagrange multiplier. The EL equations read
$$0~\approx~\frac{\delta\tilde{V}[\phi,\Lambda]}{\delta \phi^{\gamma}}~=~ -g_{\gamma\beta}(\phi)\partial^2\phi^{\beta} - \Gamma_{\gamma,\alpha\beta}(\phi)\partial_i\phi^{\alpha}\partial^i\phi^{\beta} + \Lambda ~\frac{\partial \chi(\phi)}{\partial \phi^{\gamma}}.\tag{13} $$
It is impossible to know for sure how Ref. 1 reached the wrong conclusion $V_1[\phi_1]=0$, but it might be partly spurred by forgetting to properly take into account the constraint force, i.e. the last term in eq. (13).
One-point compactification of space. In polar coordinates, the term $V_1[\phi_1]$ reads
$$ 0~\leq~2V_1[\phi_1]~=~\int_0^{2\pi} \! \mathrm{d}\theta \int_0^{\infty} \! \mathrm{d}r~\left(r \left(\frac{\partial \phi_1}{\partial r}\right)^2+ \frac{1}{r} \left(\frac{\partial \phi_1}{\partial \theta}\right)^2 \right)~<~\infty.\tag{14} $$
For each $\theta\in[0,2\pi[$, under mild regularity conditions, we can assume that the limit
$$ \lim_{r\to \infty}\left(\frac{\partial \phi_1}{\partial \theta}\right)^2\tag{15} $$
exists. To keep the energy (14) finite, the limit (15) must be zero. In other words,
$$\phi_1(r\!=\!\infty,\theta) \text{ does not depend on } \theta.\tag{16} $$
So we can effectively one-point compactify the space $\mathbb{R}^2\cup \{\infty\}\cong \mathbb{S}^2$ into a 2-sphere $\mathbb{S}^2$.
Counterexample. One counterexample with finite $V_1[\phi_1]>0$ is a so-called baby skyrmion in a $2D$ $O(3)$ model with a mexican-hat-like potential. The target space is here $\mathbb{R}^3$, and the zero locus
$$ U^{-1}(\{0\})~=~\{\phi \in \mathbb{R}^3 |~ |\phi| =1\}~\cong~\mathbb{S}^2\tag{17} $$
for the potential $U$ forms a 2-sphere. Because $\pi_2(\mathbb{S}^2)\cong \mathbb{Z}$, the field configuration $\phi_1:\mathbb{S}^2\to \mathbb{S}^2$ is protected by a topological charge
$$V_1[\phi_1]~\geq~ 4\pi |Q|.\tag{18} $$
If $Q\neq 0$, we conclude that $\phi_1$ is not a ground state. See e.g. Ref. 2 for further details.
