We have a compact and convex subset $K\subseteq \mathbb R^n$. Also assume that $K$ has more than one point.
We want to show that a point $x\in K$ is not on the boundary of $K$ (that is $x\in \operatorname{int}(K)$).
We think that it is enough to show that there exists an interval $(x-ad,x+ad)$ that is contained entirely in $K$ (that is, we can find a straight line passing through $x$ that lies in $K$). In our case $a$ is nonzero and positive and $d\notin K$.
Can we deduce from the existence of such interval the existence of an entire open neighborhood $V$ such that $x\in V \subseteq K$?
Geometrically it looks correct. But we don't have a formal proof. Any remarks would be appreciated!
Thank you!
Edit
After the discussion below, I see that in case the boundary contains a straight line, my argument is clearly false.
No, you cannot deduce such a thing. For example, for $n=2$, you can take $K=[0,1]^2$.
Then, take the point $x=(\frac12, 0)$. You can find that the line $t\mapsto (t, 0)$ for $t\in[0,1]$ (i.e. the line from $(0,0)$ and $(1,0)$) passes through $x$ and is contained entirely in $K$. Nonetheless, $x$ is not in the interior of $K$.