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.
Try the following article "A Survey of the Complemented Subspace Problem":
https://arxiv.org/abs/math/0501048v1
Your suspicion about $c_0$ is correct. A couple of other examples: The disc algebra (those functions in $C(\mathbb{T})$ which are restrictions of functions analytic in the open unit disc) is closed in $C(\mathbb{T})$ but not complemented. Similarly, in $L^1(\mathbb{T})$, the subspace $H^1(\mathbb{T})$ consisting of functions whose negative Fourier coefficients vanish is closed but not complemented. See Rudin's Functional Analysis (the proof isn't very easy).
Best Answer
There is a regular method to produce a lot of non-closed subspaces in arbitrary infinite dimensional Banach space.
Take any countable linearly independent family of vectors $\{w_i:i\in\mathbb{N}\}\subset V$ and define $W=\mathrm{span}\{w_i:i\in\mathbb{N}\}$. Then, $W$ is not closed.
Indeed, assume that $W$ is closed. Recall that $V$ is a Banach space, then $W$ is also Banach as closed subspace of Banach space. Linear dimension of $W$ is countable, but by corollary of Baire category theorem, Banach space can not have countable linear dimension. Contradiction, so $W$ is not closed.
This general result was demonstrated in Kevin's and J.J.'s answers. I'll show another one.
Consider Banach space $V=(C([0,1]),\Vert\cdot\Vert_\infty)$ of continuous functions with $\sup$ norm. Let $W=(P([0,1]),\Vert\cdot\Vert_\infty)$ be its proper subspace consisting of polynomials. It is of countable dimension because $W=\mathrm{span}\{x^k:k\in\mathbb{Z}_+\}$. From result given above it follows that $W$ is not closed.
But there is another proof. By Weierstrass theorem $W$ is dense in $V$, i.e. $\overline{W}=V\neq W$. Thus $W$ is not closed, thought it is dense in $V$.