Let $K$ be a compact convex set in a locally convex space $X$. If $U$ is a convex set contained in $K$ which is open with respect to the subspace topology in $K$ does the following hold:
If $x\in U$ and $y\in \overline{U}\setminus U$ then $(1-t)x+ty\in U$ for $0\leq t<1$. Here the closure is taken with respect to the full space $X$.
If $x\in \mathrm{int}~U$ then I know that this is true, for instance this is stated in Proposition 1.11 in Chapter IV of A course in Functional Analysis by John B. Conway. However I don't know how to prove it otherwise and would appreciate help.
I can modify Michaels argument to show that the statement is false. Let $K = [-1,1]\times [-1,1]$ which is a compact convex set and let $U = \Big((-1,1)\times(-1,1)\Big)\cup\Big((-\tfrac{1}{2},\tfrac{1}{2})\times\{1\} \Big)$. This set is open and convex in the subspace topology however take $y= (1,1)$ and $x =(0,1)$ then the open line segment between $(1,1)$ and $(0,1)$ is not contained in $U$.