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.
If $(X,\|\cdot\|_1)$ is a Banach space then $M$ is closed. Indeed, we have for $x=m+n$
$$\|x\|_1=\|m\|+\|n\|\ge \|m+n\|=\|x\|$$ Therefore the mapping $T:(X,\|\cdot\|_1)\to (X,\|\cdot\|)$ given by $Tx=x$ is a bounded bijection. By the Banach inverse mapping theorem, the inverse mapping is bounded. Hence there exists a constant $c>0,$ such that
for $x=m+n$ there holds $$\|m\|+\|n\|=\|x\|_1\le c\|x\|=c\|m+n\|\quad (*)$$
Assume $m_k\to v$ for $m_k\in M.$ Let $v=m_0+n_0.$ Then $(*)$ implies
$$\|m_k-m_0\|+\|n_0\| \le c\|(m_k-m_0)-n_0\|=c\|m_k-v\|\to 0$$ Hence $n_0=0$ and $m_k\to m_0,$ which shows that $M$ is closed. Observe that we haven't used the fact that $N$ is closed. Clearly the above reasoning implies that $N$ must be closed.
The converse implication holds as well. Indeed, assume $M$ and $N$ are closed and $X=M+N.$ Then $(X,\|x\|_1)$ is a Banach space, because if $x_k=m_k+n_k$ is a Cauchy sequence with respect to $\|\cdot\|_1$ norm then $m_k$ and $n_k$ are Cauchy sequences with respect to $\|\cdot\|.$ Hence they are convergent to $m_0$ and $n_0,$ respectively. Since $M$ and $N$ are closed, we get $m_0\in M$ and $n_0\in N.$
Best Answer
If a subsapce of a normed linear space is not dense then there is a non-zero continuous linear functional which vanishes on the subspace. Hence if $Y$ is not dense in $X^{*}$ then there exists $F \in X^{**}\setminus \{0\}$ fuch that $F=0$ on $Y$. By reflexivity $F \in X$. Do you see a contradiction to the hypotheis now?