The convex hull of a finite subset of $\Bbb{R}^2$ is the intersection of all closed half plane Let $X \subset \Bbb{R}^2$ be finite. Show that the convex hull of $X$ is the intersection of every closed half space.

This is pretty simple in a geometric way, I just can't prove it in a formal way. I've showed that the convex hull is a polygon. But this is not enough. The definition of convex hull that I'm using is the smallest convex set that contains $X$.
 A: It's pretty simple to just show this for $\mathbb R^n$, so I'll do that. Also, I don't need the finiteness condition, so I'll ignore that as well.
Let's show that a closed half space is convex. To see that the convex hull is a subset of every closed half space containing the points (and thus a subset of the intersection), observe that the intersection of convex spaces is convex.
This can be seen by observing that for any two points $x,y\in \bigcap C$, for all $X\in C$, $x,y\in X$. If all $X\in C$ are convex, then the line segment $\overline{xy}\subset X$ for all $X$, and thus $\overline{xy}\subset\bigcap C$, proving that $\bigcap C$ is convex.
Since the convex hull is the minimal convex set containing the points, this shows one direction.
For the other direction, observe that the convex hull is closed, and thus, for any $y\notin H$, the hull, $d(y,H)>0$.
Let $x\in H$ be a point such that $d(x,y)=d(H,y)$. Such a point exists because the hull is closed. Observe that there can be no point other than $x$ in the line segment $\overline{xy}$ in $H$, since the distance of point $tx+(1-t)y$ for $t\in[0,1)$ is $td(x,y)<d(x,y)$.
Consider a hyperplane $A$ normal to the vector $\vec{xy}$, and some constant $b$ such that $Az=b$ has a solution in $\{tx+(1-t)y\mid t\in(0,1)\}$. Note that for all points in the set of points that generate the hull, they all either satisfy $Az\leq b$, or $Az\geq b$. Assume WLOG $Az\leq b$ (negate $A,b$ otherwise). Thus, the entire convex hull must satisfy $Az\leq b$. But, $Ay>b$ by construction.
