Making my comments into an answer: No there are no such Banach spaces.
Assume that every proper subspace of $X$ is reflexive. Take a non-zero continuous linear functional $\varphi: X \to \mathbb{R}$. Let $Y = \operatorname{Ker}{\varphi}$ and choose $x_0 \in X$ with $\varphi(x_0) = 1$. By continuity of $\varphi$ the space $Y$ is a closed subspace. The map $Y \oplus \mathbb{R} \to X$ given by $(y,t) \mapsto y + t x_0$ is continuous with continuous inverse $x \mapsto (x - \varphi(x)\cdot x_0,\, \varphi(x))$, hence $X \cong Y \oplus \mathbb{R}$. By hypothesis $Y$ is reflexive, hence so is $Y \oplus \mathbb{R}$ and thus $X$ is reflexive, too. Replacing $\mathbb{R}$ by $\mathbb{C}$ gives the same for complex Banach spaces.
Weakening the hypotheses as Robert suggested makes it a bit more subtle, but still manageable:
Every non-reflexive Banach space contains a non-reflexive closed subspace of infinite codimension.
(Bourbaki, Topological vector spaces 1, Exercise 12 to Chapter IV, §5, page IV.69.)
Passing to the contrapositive, if every closed subspace of infinite codimension is reflexive, then $X$ must be reflexive.
The following is a slightly expanded version of the hint given by Bourbaki:
By the Eberlein–Šmulian theorem, a non-reflexive Banach space $X$ contains a bounded sequence $(x_n)_{n=1}^{\infty}$ without weak accumulation point. Note that this implies that the $x_n$ must span an infinite-dimensional subspace of $X$. Using the Riesz-lemma on almost orthogonal vectors to closed subspaces, it is not difficult to extract a subsequence $(x_{n_k})_{k=1}^{\infty}$ and a topologically independent sequence $(y_k)_{k=1}^{\infty}$ such that $\|x_{n_k} - y_{k}\| \leq \frac{1}{k}$ [topologically independent means that no $y_k$ is in the closed linear span of $\{y_n\}_{n \neq k}$ ]. This yields that every weak accumulation point of the $y_k$ is also a weak accumulation point of the $x_n$. The closed subspace $Y$ generated by the $\{y_{2k}\}$ is thus a non-reflexive subspace of $X$ (by Eberlein–Šmulian again) and it has infinite codimension by construction.
From
http://www.proofwiki.org/wiki/Banach-Alaoglu_Theorem
there were just a few more steps:
- Let $l(g_i) = \lim_{n_k} \langle x_{n_k} , g_i \rangle = \langle x_\infty, g_i \rangle$. (We can obtain $x_\infty$ by including the standard basis in the dense $g_i$).
- Extend to all $g \in l^2$ via $l(g) = \lim_i l(g_i)$. This makes $l$ a linear functional, and can be bounded based on $x_{n_k}$. $|l(g)| \leq |l(g)-l(g_i)| + |l(g_i) - l_{n_k}(g_i)| + |l_{n_k}(g_i)|$.
- Riesz representation and uniqueness shows that $l(x) = \langle x_\infty, x \rangle$.
Then we have the weak convergence as desired.
Best Answer
The unit ball in the dual space $X^{*}$ is always compact under the weak$^{*}$ topology by the Banach-Alaoglu theorem, however, recall that for a general topological space compactness and sequential compactness are not equivalent. If X is separable one can show (see for example Theorem 3.28 in Haim Brezis book) that the unit ball in the dual space is metrizable and hence also sequentially compact.
One can sometimes give an explicit characterization of the dual space, as in the case of $L^{p}$, for $1 < p < \infty$. There is also a result that says if $Y$ is a closed subspace of a reflexive space $X$, then $Y$ is reflexive.
I have no useful comments here.