[proofplan]
We first separate the degenerate case in which the infimum is $+\infty$, where every admissible point is already a minimizer. In the finite case, we choose a minimizing sequence and use coercivity to prove that it is bounded in $X$. Reflexivity gives a weakly convergent subsequence, sequential weak closedness keeps the weak limit inside $\mathcal A$, and sequential weak lower semicontinuity passes the minimizing inequality to the limit.
[/proofplan]
[step:Reduce to the case where the infimum is finite]
Define
\begin{align*}
m:=\inf_{u\in\mathcal A} I[u]\in[-\infty,+\infty].
\end{align*}
Since $I$ is bounded below on $\mathcal A$, we have $m>-\infty$.
If $m=+\infty$, then $I[u]=+\infty$ for every $u\in\mathcal A$. Since $\mathcal A$ is nonempty, choose $u_*\in\mathcal A$. Then
\begin{align*}
I[u_*]=+\infty=m,
\end{align*}
so $u_*$ attains the infimum.
It remains to prove the result under the assumption
\begin{align*}
-\infty<m<+\infty.
\end{align*}
[/step]
[step:Choose a minimizing sequence with controlled energy]
For each $k\in\mathbb N$, the definition of $m$ as an infimum and the finiteness of $m$ give an element $u_k\in\mathcal A$ such that
\begin{align*}
m\le I[u_k]\le m+\frac{1}{k}.
\end{align*}
Thus $(u_k)_{k=1}^{\infty}$ is a minimizing sequence in $\mathcal A$, meaning
\begin{align*}
\lim_{k\to\infty} I[u_k]=m.
\end{align*}
In particular, the sequence of values $(I[u_k])_{k=1}^{\infty}$ is bounded above.
[/step]
[step:Use coercivity to prove the minimizing sequence is bounded]
We claim that $(u_k)_{k=1}^{\infty}$ is bounded in $X$. Suppose not. Then there is a subsequence $(u_{k_j})_{j=1}^{\infty}$ such that
\begin{align*}
\|u_{k_j}\|_X\to\infty.
\end{align*}
Since each $u_{k_j}\in\mathcal A$ and $I$ is coercive on $\mathcal A$, it follows that
\begin{align*}
I[u_{k_j}]\to+\infty.
\end{align*}
This contradicts the upper bound
\begin{align*}
I[u_{k_j}]\le m+\frac{1}{k_j}\le m+1.
\end{align*}
Therefore there exists a constant $R>0$ such that
\begin{align*}
\|u_k\|_X\le R
\end{align*}
for every $k\in\mathbb N$.
[guided]
The role of coercivity is exactly to prevent minimizing sequences from escaping to infinity in norm. We prove this by contradiction.
Assume that the minimizing sequence $(u_k)_{k=1}^{\infty}$ is not bounded in $X$. By the definition of boundedness in a normed space, this means that for every $R>0$ there exists an index $k\in\mathbb N$ with $\|u_k\|_X>R$. Hence we may choose a subsequence $(u_{k_j})_{j=1}^{\infty}$ such that
\begin{align*}
\|u_{k_j}\|_X\to\infty.
\end{align*}
Now we use the definition of coercivity on $\mathcal A$. The subsequence still lies in $\mathcal A$, because every term of the original minimizing sequence lies in $\mathcal A$. Since its $X$-norm tends to infinity, coercivity gives
\begin{align*}
I[u_{k_j}]\to+\infty.
\end{align*}
But this is incompatible with the way the minimizing sequence was chosen. For every $j\in\mathbb N$, we have
\begin{align*}
I[u_{k_j}]\le m+\frac{1}{k_j}.
\end{align*}
Since $k_j\ge 1$, this implies
\begin{align*}
I[u_{k_j}]\le m+1.
\end{align*}
Thus the same sequence of real extended values is both bounded above by the finite number $m+1$ and tends to $+\infty$, a contradiction. Therefore the original sequence must be bounded in $X$.
[/guided]
[/step]
[step:Extract a weakly convergent subsequence by reflexivity]
Since $X$ is reflexive and $(u_k)_{k=1}^{\infty}$ is bounded in $X$, the [weak sequential compactness](/theorems/214) theorem for bounded sequences in reflexive Banach spaces gives a subsequence $(u_{k_j})_{j=1}^{\infty}$ and an element $u_*\in X$ such that
\begin{align*}
u_{k_j}\rightharpoonup u_*
\end{align*}
weakly in $X$.
[/step]
[step:Use weak closedness to keep the limit admissible]
The subsequence $(u_{k_j})_{j=1}^{\infty}$ lies in $\mathcal A$ and converges weakly in $X$ to $u_*$. Since $\mathcal A$ is sequentially weakly closed, the weak limit belongs to $\mathcal A$. Hence
\begin{align*}
u_*\in\mathcal A.
\end{align*}
[/step]
[step:Apply weak lower semicontinuity to identify the minimizer]
Because $I$ is sequentially weakly lower semicontinuous on $\mathcal A$, because $u_{k_j}\in\mathcal A$, and because $u_{k_j}\rightharpoonup u_*$ weakly in $X$, we have
\begin{align*}
I[u_*]\le \liminf_{j\to\infty} I[u_{k_j}].
\end{align*}
The sequence $(I[u_k])_{k=1}^{\infty}$ converges to $m$, so every subsequence of it also converges to $m$. Therefore
\begin{align*}
\liminf_{j\to\infty} I[u_{k_j}]=m.
\end{align*}
Thus
\begin{align*}
I[u_*]\le m.
\end{align*}
On the other hand, since $u_*\in\mathcal A$ and $m$ is the infimum of $I$ over $\mathcal A$, we have
\begin{align*}
m\le I[u_*].
\end{align*}
Combining the two inequalities gives
\begin{align*}
I[u_*]=m=\inf_{u\in\mathcal A} I[u].
\end{align*}
Thus $I$ attains its infimum on $\mathcal A$.
[/step]
