Let $X$ be a Banach space with continuous dual space $X'$ with norm topology. Let us regard the following property of $X$:
Property: Any linear subset $A \subset X'$ that satisfies $\bigcap_{\alpha\in A} \ker\alpha = \{0\}$ is dense in $X'$.
Any reflexive space $X$ has this property. Can you classify the spaces that share this property? I wonder whether it is equivalent to reflexivity.
Let $\Phi\in X''$ be non-zero, let $\alpha_0\in X'$ be such that $\Phi(\alpha_0)=1$, and let $A$ be the collection of $\alpha\in X'$ with $\Phi(\alpha)=0$. Then $A$ is a subspace of $X'$, and any $\beta\in X'$ is equal to
\[ \beta = \Phi(\beta)\alpha_0 + (\beta - \Phi(\beta)\alpha_0) \in \mathbb K\alpha_0 \oplus A, \]
where $\mathbb K\alpha_0$ means the span of single element $\alpha_0$ (over whatever field you are using). In particular, $A$ is not norm dense.
Suppose now that $x\in X$ is non-zero, but with $\alpha(x)=0$ for all $\alpha\in A$. Then for any $\beta\in X'$ we have that
\[ \beta(x) = \Phi(\beta)\alpha_0(x) + (\beta - \Phi(\beta)\alpha_0)(x)
= \Phi(\beta) \alpha_0(x). \]
Thus $x = \alpha_0(x) \Phi$, in particular, as $x$ is non-zero, $\alpha_0(x)\not=0$, and so $\Phi = \alpha_0(x)^{-1} x$.
Thus, if $X$ is not reflexive, I can choose $\Phi\in X''\setminus X$, and then I have found a suitable set $A$ which is not dense. So your condition is equivalent to reflexivity.
