You are saying that if a collection of planar sets does have a non-empty intersection it is visually obvious so what use is a theorem asserting the fact (at least if they are convex and we can see them all at the same time?)

The interesting thing is not the examples, it is the lack of counter-examples. Imagine four or five long thin ovals all sharing a common point. Can you move them around a bit so that no three have empty intersection yet the whole family does? It seems not, there always seems to be a common point, but how would you prove it? That is Helley's theorem.

Here is a proof of Let $M$ be a finite set of points in the plane, with all pairwise distance between them not exceeding $1$. Then $M$ is contained in a disk of radius $\frac{1}{\sqrt3}$. See if you can figure out the short proof. So again in any specific case perhaps we can find an appropriate circle but we need the theorem to say that we won't find a counter-example.

This seems more contrived, but I can attack "at the same time." imagine that I give you $\binom53=10$ cards each with a picture of three colored ovals ( all possible combinations of red, blue , green , yellow and black. ) I tell you that these are the same five sets, viewed three at a time. My claim looks reasonable and you can see that every triple has a non-empty intersection. It might not be obvious where a common point of all five is, but if I am telling the truth then there must be one.

You are saying that if a collection of planar sets does have a non-empty intersection it is visually obvious so what use is a theorem asserting the fact (at least if they are convex and we can see them all at the same time?)

The interesting thing is not the examples, it is the lack of counter-examples. It is easy to draw $n$ convex sets in the plane so that no two are disjoint yet no point is common to all of them. Can you change two to three? It does not seem like it but that is no proof. You can change to three for bodies in $\mathbb{R}^3$ but not four. The problem is understandable for $\mathbb{R}^m$ but geometric intuition may be weaker.

Here is a proof of Let $M$ be a finite set of points in the plane, with all pairwise distance between them not exceeding $1$. Then $M$ is contained in a disk of radius $\frac{1}{\sqrt3}$. See if you can figure out the short proof. So again in any specific case perhaps we can find an appropriate circle but we need the theorem to say that we won't find a counter-example.

This seems more contrived, but I can attack "at the same time." imagine that I give you $\binom53=10$ cards each with a picture of three colored ovals ( all possible combinations of red, blue , green , yellow and black. ) I tell you that these are the same five sets, viewed three at a time. My claim looks reasonable and you can see that every triple has a non-empty intersection. It might not be obvious where a common point of all five is, but if I am telling the truth then there must be one.