Example of Three Polyhedrons in $\mathbb {R^2}$ where any Two Have Non-Empty Intersection, but Empty Intersection Amongst All Suppose we are in $\mathbb R^2$.
It is not true that the polyhedra $P_1, P_2$,...$P_K$ (where $K \geq 3$) in the plane have a point in common if every two of them have a point in common.

For example:
Let $P_1=\{(x,y) \in \mathbb R^2 \mid x=0\}$, $P_2=\{(x,y) \in \mathbb {R^2} \mid y=0 \}$, $P_3=\{(x,y) \in \mathbb {R^2} \mid y=x-1 \}$

Any two have a point in common but the intersection amongst all three is empty. This is an interesting and cool example.
I was wondering what some other possible examples of polyhedron were to illustrate this fact.
In fact, if you want bounded polygons, it suffices to "bend" in a suitable way the straight lines solution ; for example for $K=3$ :



This method is suitable in any dimension ; consider for example the case of $K=5$ unbounded regions delimited by $5$ straight lines :



Their boundaries can be "bended" into closed polygons, giving this solution:
Let $P_1 = \{ (x,y) \in \mathbb{R}^2| x \le 0 \}$, $P_2 = \{ (x,y) \in \mathbb{R}^2 | y \le 0\}$, $P_3 = \{(x,y): x+y \ge 1 \}$