I will give an example-based answer to this and I hope I will provide some intuition of the idea in the end.
Let $X$ be a uniformly convex Banach space, and $C\subset X$ be a nonempty, closed, convex subset.
Let $x_0 \in X$ be fixed. We consider the projection problem:
$$ \text{min} f(x) = ||x-x_0||$$
over $x\in C$, and we denote the infimum $r:= \text{inf}_{x \in C} ||x-x_0|| $.
Let $(x_n) \subset C$ be such that
$$ ||x_n-x_0|| \to r$$
It is trivial to show that $(x_n) $ is bounded. Hence by Milman–Pettis, Banach-Alaoglu etc, it has a weakly convergent subsequence i.e.
$$ x_{n_k} \rightharpoonup x^*$$
We will prove $x^* \in C$ and that it is the minimizer:
The norm in any Banach space is weakly lower semicontinuous:
$$ ||x^* - x_0|| \leq \text{liminf}_{k \to \infty} ||x_{n_k}-x_0||$$
So
$$ ||x^*-x_0|| \leq r$$
But since $r$ is the infimum we must have
$$||x^*-x_0|| = r$$
so indeed $x^*$ is a minimizer and the infimum is attained.
One can actually prove the minimizing sequence strongly converges to the minimizer.
In a general Banach space, minimizing sequences might not converge. They might move around weakly without converging strongly. But in a uniformly convex space, the geometry forbids this and we'll see why:
Suppose $(x_n)$ and $(y_n)$ are two minimizing sequences, i.e.
$$||x_n-x_0|| \to r$$ and $$||y_n-x_0|| \to r$$
From uniform convexity we have that for every $\epsilon >0$ there exists some $\delta(\epsilon)>0$ such that:
$$ ||x||=||y||=1, ||x-y|| \geq \epsilon \Rightarrow ||\dfrac{x+y}{2}|| \leq 1-\delta(\epsilon)$$
We can normalize as
$$u_n = \dfrac{x_n-x_0}{r}, υ_n = \dfrac{y_n-x_0}{r}$$
Then $||u_n||, ||υ_n|| \to 1$. Now if $||x_n-y_n|| \nrightarrow 0$ then there exists an $\epsilon >0$ such that for infinitely many $n$
$$||u_n-υ_n|| \geq \epsilon$$
By uniform convexity
$$|| \dfrac{u_n+υ_n}{2}|| \leq 1-\delta(\epsilon)$$
Scaling back we have
$$||\dfrac{x_n+y_n}{2}-x_0|| = r||\dfrac{u_n+υ_n}{2}|| \leq r(1-\delta(\epsilon))$$
But this is a contradiction since the point $\dfrac{x_n+y_n}{2} \in C$ because $C$ is convex, and from the above for large $n$
$$||x_0 - \dfrac{x_n+y_n}{2}|| <r$$
But $r$ was the infimum distance so necessarily we have $||x_n-y_n|| \to 0$.
Geometric intuition of uniform convexity:
What the above says is that no minimizing sequence can "escape" not even weakly because any two minimizing sequences get arbitrarily close. This happens because uniform convexity gives the unit ball a "uniformly curved" boundary, so any almost minimizing point is geometrically forced inward. This prevents the "flat escape directions" that non reflexive spaces have. As a result the ball is weakly compact and no sequence can drift away without a weak limit. Think of it like the curvature making the "valley" of any convex functional strictly "bowl" shaped, so minimizing sequences cannot escape.
Disclaimer:
This is my best understanding of why uniform convexity implies reflexivity, viewed through the minimization of functionals. I’m not entirely sure whether my analysis is helpful, but it makes some intuitive sense to me, so that's why I'm posting it here. Any corrections or additions are always welcome.
