Show that the set $U$ contains the set $S$
Solution 1:
Hint
$U$ and $S$ are connected (and convex) subsets of the plane. To prove that $S \subseteq U$, it is enough to prove that:
- One point of $S$, i.e. $(1,0)$ belongs to $U$.
- And that the boundary of $S$ doesn't intersect the boundary of $U$.
The boundary of $U$ is the circle centered on the origin with radius $\sqrt 3$. And the boundary of $S$ is a portion of a circle plus two line segments.