Let $S$ be a convex set. If $x\in$ int$S$ and $y\in$ cl$S$, show that relint[x,y] $\subset$ int$S$.
I easily proved this for a case where y is in the interior of S, but am stuck if y is in the boundary. S need not be closed by assumption, so I either need to contradict that or need to show that y is in S, making S closed.
Here are some broad strokes. I think they'll lead to a complete proof rather directly. But please let me know if I'm wrong.
Since $S$ is a convex set, its closure is a convex set.
Since $cl(S)$ is a convex set, the line segment containing any two points in the set lies in $cl(S)$.
Since $x\in \text{int}(S)$, at least part of the line segment connecting x and y lies in $\text{int}(S)$.
If you suppose that a point on $\text{relint}[x,y]$ other than $y$ lies on the boundary of S, this leads to a contradiction.
Since $y$ lies on the boundary of $S$ and $x$ lies in the interior of $S$, all points on $\text{relint}[x,y]$ lie in the interior of $S$.