Suppose $X$ is a finite-dimensional normed space over $\mathbb{R}$. Let $\emptyset\neq M\subseteq X$ be a convex set with empty interior, and let $x\in\overline{M}\backslash M$ be a point on the boundary that is disjoint from $M$. I would like to expand $M$ to a convex set with nonempty interior that is still disjoint from $x$, and the method I thought of is to prove the existence of some open ball $B$ such that the convex hull of $B\cup M$ will remain disjoint from $x$ (or alternatively find a counterexample).
I have assumed WLOG that $0\in M$. Note that therefore $\text{span}M\neq X$, as otherwise $M$ would contain an $n$-simplex ($n$ being the dimension of $X$), contradicting the fact that $M$ has empty interior.
It seems to me that this is in fact possible, at least from trying to picture it in my head or on paper, yet I struggle to formalize this into a proof. Any help will be greatly appreciated.
EDIT: I thought of this problem in the context of solving an exercise in functional analysis where we must find a linear functional separating $M$ from $x$, thereby extending the Hahn-Banach separation theorem to this case where neither set has an internal point. Therefore, I would like to refrain from using a result of a generalized separation theorem, as that is ultimately what I am trying to show (I came up with an alternate solution to the exercise, but the question here seemed interesting to me on it's own).
I believe I have found a solution. By restricting momentarily to the subspace $\text{span}M$, $M$ has nonempty interior in this subspace, so I can take a point $y \in M$ and find an open ball (in $X$) $B:=B\left(y,r\right)$ such that $B\cap\text{span}M\subseteq M$. The convex hull of $B \cup M$ is all the convex combinations $tb+\left(1-t\right)m$ for $b \in B$ and $m \in M$, since both sets are convex.
All that remains is to show that such a convex combination can never give $x$. If $b$ lies in the intersection $B\cap M$ or $t=0$ then obviously the combination will remain in $M$. Otherwise the combination will give a point that lies outside of $\text{span}M$, but $\text{span}M$ is a closed set which contains $M$ and therefore contains $x$.
EDIT: I should note that since $M$ is not closed, it cannot be the singleton $\left\{ 0\right\} $, which lets me assume that $\text{span}M$ is nontrivial.