Expanding a convex set to a convex set with nonempty interior, while maintaining disjointness from a point

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.