In additional to the standard definition of "the interior of a set" in a topological space, Kolmogorov & Fomin define (sec. 14.1) another concept of "interior" in a real vector space that does not require a topology: the interior $I(E)$ of a set $E$ in a vector space is the set of all such $x \in E$ that, for every vector $y$ there exists a positive $\epsilon_{y}$ (generally dependent on $y$, so a weaker requirement than "open neighborhood") all the vectors $x + ty, \; |t| < \epsilon$, lie in $E$.
Intuitively, this latter concept of interior (henceforth, non-topological interior) is the set of all points from which, no matter what direction you go, you must go a "positive distance" before you can leave $E$. This distance depends on the direction, so the concept is weaker than the standard topological interior.
So, in a real topological vector space, the (standard topological) interior of a set $E$ is contained in the non-topological interior $I(E)$. I am looking for examples of sets $E$ with an empty interior, but nonempty $I(E)$.
If $E$ is a Banach and $0 \in C \subset E$ is a closed convex subset with $0 \in I(C)$ then $C$ has nonempty interior (in fact, $0$ is adherent to the interior of $C$).
Proof: let $S \subset E$ be the unit sphere, and $U_n=\{y \in S,\, y/n \in S\}$. Then each $U_n$ is a closed subset of $S$, and $S=\cup_n{U_n}$. So by Baire, there exist some $n \geq 1$, $V \subset S$ nonempty open, such that $V \subset U_n$. Then $(0,1/n)V \subset C$ is a nonempty open subset whose closure contains $0$.
So, if we want to try and find closed convex $C$ with empty interiors but $I(C) \neq \emptyset$, we need to consider non-complete spaces.
Let $E$ be the space of real sequences with finitely many nonzero terms, with the uniform norm (ie $\|u\|=\max_n\,|u_n|$).
Take $C=\{u \in E|\, \forall n,\, |nu_n| \leq 1\}$. Then you can easily check that $0 \in I(C)$ while $C$ has empty interior.