Based on Jack's comment.
The "Hilbert cube" in Hilbert space $l^2$. $$C :=\{(x_1,x_2,\dots) : |x_k| \le 2^{-k}\;\forall k\}$$ $C$ is convex, compact (so it has empty interior) but has dense span (so it not contained in a closed hyperplane).
However, $C$ is contained in a (non-closed) hyperplane. (Axiom of Choice required.)
The linear span of $C$ is not the whole of $l^2$. Indeed, if $(a_1,a_2,\dots)$ is in the span of $C$, then
$\limsup_k 2^k |a_k| < \infty$, but $(1,1/2,1/3,\cdots) \in l^2$ fails that property.
Now we define a Hamel basis for $l^2$ as follows: first choose any vector $u$ not in the span of $C$; next add to {u} a Hamel basis of the span of $C$; then extend that to a Hamel basis $B$ of $l^2$. Then we can see that $C$ is contained in the hyperplane
$$
\left\{\sum_{b \in B} t_b b : t_u = 0\right\}
$$