Below is my explicit example of a uniformly convex normed vector space which is not reflexive. Could you confirm if my understanding is correct?
Let $$ \ell_2 := \big \{ (v_n)_{n \ge 1} \subset \mathbb R^\mathbb N : \sum_n |v_n|^2 < \infty \big \} $$ be equipped with the canonical inner product $\langle u, v \rangle := \sum_n u_n v_n$. Then $(\ell_2, \langle \cdot, \cdot \rangle)$ is a separable Hilbert space. Let $$ V := \big \{ (v_n)_{n \ge 1} \subset \mathbb R^\mathbb N : \sum_n n^2 |v_n|^2 < \infty \big \}. $$
Then $V$ is a vector subspace of $\ell_2$. We have $(1, 1/2, 1/3, \ldots) \in \ell_2 \setminus V$, so $V \subsetneq \ell^2$. We define $e^m \in \ell^2$ by $e^m_n = 1$ if $n=m$ and $0$ otherwise. Then $(e^m)$ is an orthogonal basis of $\ell^2$. Clearly, $(e^m) \subset V$, so $V$ is dense in $\ell_2$. Hence $(V, \langle \cdot, \cdot \rangle)$ is an incomplete inner product space.
We define a new inner product $\langle \cdot, \cdot \rangle_V$ on $V$ by $\langle u, v \rangle_V := \sum_n n^2 u_n v_n$. Let's prove that $(V, \langle \cdot, \cdot \rangle_V)$ is a Hilbert space. We define a $\sigma$-finite measure $\mu$ on $\mathbb N$ by $\mu(n) := n^2$. Clearly, every $v \in \mathbb R^\mathbb N$ is $\mu$-measurable. For $v \in \mathbb R^\mathbb N$, we have $$ \| v \|_{L^2(\mathbb N, \mu, \mathbb R)} = \sqrt{\sum_n n^2 |v_n|^2} = \sqrt{\langle v, v \rangle_V}. $$
Then $L^2(\mathbb N, \mu, \mathbb R) = V$. It follows from $L^2(\mathbb N, \mu, \mathbb R)$ is complete that $(V, \langle \cdot, \cdot \rangle_V)$ is complete.