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.
Best Answer
Reflexive spaces interest mathematicians because they have a lot of nice properties:
Unit ball is weakly compact, so you can exploit compactness to prove exitence of fixed points, convergent subsequences and etc.
Reflexive spaces are characterized by the property that weak and weak* topology coincide. You can forget about weak topology and work with much well understood weak* topology.
Every functional on a reflexive space attains its norm. Simply speaking you always have a vector that tells you almost everything about your functional. More on this matter you can find here.
Reflexivity is a three space property. You can pass to quotients and subspaces of reflexive spaces and get a reflexive space again.
After equivalent renorming all reflexive spaces are strictly convex. In some sense unit ball of a reflexive spaces is round.
Reflexive spaces have Radon-Nykodim property. This allows you to develop a rich theory for vector valued integration and vector valued measures for reflexive spaces.
Reflexivity is a rare property and this helps one to distinguish Banach spaces. For example there is no infinite dimensional reflexive $C^*$-algebras, so $c_0$, $l_\infty$ are not reflexive. Their non-commutative counterparts $\mathcal{K}(H)$ and $\mathcal{B}(H)$ are not reflexive either.
Shauder bases in a reflexive space are very nice and sweet, they are shrinking and boundedly complete.. Beware! There are hereditarily indecomposable (and a fortiori without any basis) reflexive Banach spaces. See this paper.
To find more on reflexive spaces use search on this site, or mathoverflow, or any book on Banach geometry with keyword 'reflexive'.