First, as
$$
\Delta u(x) = \phi (x)
$$
we have
$$
0 = \int_U d^nx \left( - \Delta u(x) + \phi (x) \right) \eta (x)
$$
for an arbitrary function ##\eta (x)##. If we take ##\eta \in \Sigma##, integrating by parts gives
$$
0 = \int_U d^nx \left( \nabla u(x) \cdot \nabla \eta (x)+ \phi (x) \eta (x) \right)
$$
Take an arbitrary ##v \in \Sigma##. We can write down an obvious inequality
$$
0 \leq \left( \| \nabla u \| - \| \nabla v \| \right)^2
$$
where ##\| \nabla u \| := \left( \sum_{i=1}^n (\partial_i u)^2 \right)^{1/2}## (##\| \nabla u \|^2 \equiv | \nabla u |^2##). Which implies
$$
0 \leq \frac{1}{2} | \nabla u |^2 + \frac{1}{2} | \nabla v |^2 - \| \nabla u \| \| \nabla v \|
$$
or
$$
\| \nabla u \| \| \nabla v \| \leq \frac{1}{2} | \nabla u |^2 + \frac{1}{2} | \nabla v |^2 .
$$
As ##\nabla u \cdot \nabla v \leq | \nabla u \cdot \nabla v | \leq \| \nabla u \| \| \nabla v \|## (Cauchy-Schwarz) we have
$$
\nabla u \cdot \nabla v \leq \frac{1}{2} | \nabla u |^2 + \frac{1}{2} | \nabla v |^2 .
$$
From which we have
$$
\int_U d^nx \left( \nabla u \cdot \nabla v + \phi v - \frac{1}{2} | \nabla u |^2 \right) \leq \int_U d^nx \left( \frac{1}{2} | \nabla v |^2 + \phi v \right)
$$
Using ##\int_U d^nx \left( \nabla u \cdot \nabla \eta+ \phi \eta \right) = 0## in this integral gives
\begin{align*}
\int_U d^nx \left( \nabla u \cdot \nabla [\eta + v] + \phi [\eta + v] - \frac{1}{2} | \nabla u |^2 \right)
\leq \int_U d^nx \left( \frac{1}{2} | \nabla v |^2 + \phi v \right)
\end{align*}
Putting ##\eta (x) = u(x) - v(x)## we have,
$$
\int_U d^nx \left( \frac{1}{2} | \nabla u (x) |^2 + \phi (x) u (x) \right) \leq \int_U d^nx \left( \frac{1}{2} | \nabla v (x) |^2 + \phi (x) v (x) \right)
$$
which is the desired result. Obviously, this choice of ##\eta (x)## means that ##\eta \in \Sigma##.