To show uniqueness is easy, and not covered by other answers. If you have two distinct convex polygons which contain all your points, then their intersection is also a convex polygon which contains all the points.
The two original polygons cannot both be minimal unless they coincide.
To use your rubber band analogy, assuming $P$ is finite, you could proceed as follows.
First contain your points in a finite square with horizontal and vertical sides. This confines your points in a finite convex set. Note also that a line divides the plane into two half-planes - these half planes are convex, and the intersection of two convex sets is convex.
Now identify the top point of your set (or one of them). Draw a horizontal line through this point, so all the points not on the line are below it. Call the leftmost of the points (perhaps there is only one) $P_1$. Now rotate the line clockwise about $P_1$ until it meets another point (it may meet more than one). This becomes $L_1$ and the point on $L_1$ furthest from $P_1$ we call $P_2$ - if we are facing from $P_1$ to $P_2$ all the points are on our right. We reduce the square by intersecting it with this right half-plane. We then rotate about $P_2$ to find $L_2$ and $P_3$ etc, always keeping all the points on our right and cutting off parts of the original square as we go.
Since we have only a finite number of points, we can't keep going for ever. When the line comes horizontal again, with all the points on the right i.e. below, it must go through $P_1$ otherwise $P_1$ would be above the line.
I think that can be made rigorous.