Both facts have short proofs. Let's prove (1).

**Proof of (1)**. By assumptions, for each point $x$ of the set there is a half-space $H_x$ containing  $x$ in its boundary and containing all other points in its interior. So there is also a radius $R$ ball $B_x^R$ contained in $H_x$, tangent to $\partial H_x$ at $x$ and containing all the other points of the set (just take $R$ large enough). Now, the intersection of all balls $B_x^R$ is the desired strictly convex set.

**Proof of (2).** Same idea, but use sets $y\ge f(x)$ where $f$ is a quadratic polynomial instead of round balls.