[proofplan]
Restrict the functional $I$ to the admissible affine line through $u_*$ in the direction $v$. The minimality of $u_*$ over $\mathcal A$ implies that this one-variable restriction has a minimum at the interior point $0$. Since the restriction is differentiable at $0$, its right and left difference quotients have the same limit, while minimality forces them to have opposite signs. Therefore the derivative at $0$ is zero, and this derivative is exactly the first variation $\delta I[u_*;v]$.
[/proofplan]
[step:Restrict the functional to the admissible variation line]
Define the function
\begin{align*}
\phi:(-\varepsilon_0,\varepsilon_0)\to\mathbb R,\qquad \phi(\varepsilon)=I[u_*+\varepsilon v].
\end{align*}
This is well-defined because $u_*+\varepsilon v\in\mathcal A$ for every $\varepsilon\in(-\varepsilon_0,\varepsilon_0)$, and $I$ is defined on $\mathcal A$. Since $u_*+0v=u_*$, we have
\begin{align*}
\phi(0)=I[u_*].
\end{align*}
[/step]
[step:Use minimality to show that $\phi$ has an interior minimum at $0$]
Let $\varepsilon\in(-\varepsilon_0,\varepsilon_0)$. By admissibility, $u_*+\varepsilon v\in\mathcal A$. Since $u_*$ minimizes $I$ over $\mathcal A$, we get
\begin{align*}
I[u_*]\le I[u_*+\varepsilon v].
\end{align*}
Using the definition of $\phi$, this becomes
\begin{align*}
\phi(0)\le \phi(\varepsilon).
\end{align*}
Thus $\phi$ has a minimum at $0$ relative to the open interval $(-\varepsilon_0,\varepsilon_0)$. The point $0$ is an interior point of this interval because $\varepsilon_0>0$.
[guided]
The purpose of introducing $\phi$ is to turn the variational statement into a one-variable calculus statement. The hypothesis says that every point of the form $u_*+\varepsilon v$ remains admissible as long as $\varepsilon\in(-\varepsilon_0,\varepsilon_0)$. Therefore the function
\begin{align*}
\phi:(-\varepsilon_0,\varepsilon_0)\to\mathbb R,\qquad \phi(\varepsilon)=I[u_*+\varepsilon v]
\end{align*}
is a genuine real-valued function on an open interval.
Now fix any $\varepsilon\in(-\varepsilon_0,\varepsilon_0)$. The point $u_*+\varepsilon v$ belongs to $\mathcal A$, so the global minimality assumption on $u_*$ applies to this point. Hence
\begin{align*}
I[u_*]\le I[u_*+\varepsilon v].
\end{align*}
Since $\phi(0)=I[u_*]$ and $\phi(\varepsilon)=I[u_*+\varepsilon v]$, the inequality is exactly
\begin{align*}
\phi(0)\le \phi(\varepsilon).
\end{align*}
This holds for every $\varepsilon$ in the open interval. Thus $0$ is a minimum point of the one-variable function $\phi$. The openness of the interval matters: because $\varepsilon_0>0$, both positive and negative perturbations of $0$ are available, which is what will force the derivative to vanish rather than merely have one sign.
[/guided]
[/step]
[step:Compare right and left difference quotients at the minimum]
For every $h\in(0,\varepsilon_0)$, the inequality $\phi(0)\le\phi(h)$ gives
\begin{align*}
\frac{\phi(h)-\phi(0)}{h}\ge 0.
\end{align*}
For every $h\in(-\varepsilon_0,0)$, the same inequality $\phi(0)\le\phi(h)$ and the negativity of $h$ give
\begin{align*}
\frac{\phi(h)-\phi(0)}{h}\le 0.
\end{align*}
Since $\phi$ is differentiable at $0$, the two-sided limit
\begin{align*}
\phi'(0)=\lim_{h\to 0}\frac{\phi(h)-\phi(0)}{h}
\end{align*}
exists. The right-hand difference quotients have nonnegative limit, so $\phi'(0)\ge 0$. The left-hand difference quotients have nonpositive limit, so $\phi'(0)\le 0$. Therefore
\begin{align*}
\phi'(0)=0.
\end{align*}
[/step]
[step:Identify the derivative with the first variation]
By the definition of the first variation in the direction $v$,
\begin{align*}
\delta I[u_*;v]=\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0} I[u_*+\varepsilon v].
\end{align*}
The right-hand side is precisely $\phi'(0)$. Since $\phi'(0)=0$, we conclude that
\begin{align*}
\delta I[u_*;v]=0.
\end{align*}
This proves the theorem.
[/step]
