How to understand case by case decomposition in combinatorics proof I am just reading through Mathematics for Computer Science (https://courses.csail.mit.edu/6.042/spring17/mcs.pdf) and at chapter 1.7 Proof by Cases there is a following theorem as an example:

Theorem. Every collection of 6 people includes a club of 3 people or a
  group of 3 strangers.

The author(s) state the following:

Proof. The proof is by case analysis
  Let x denote one of the six
  people. There are two cases:
  1. Among 5 other people besides x, at least 3 have met x.
  2. Among the 5 other people, at least 3 have not met x

The chapter goes on splitting the cases in the 4 subcases (2 for each case). I just can not see how proving these 2 cases proves the theorem. How did stating that 3 other people have met or not met the fourth (making it a group of 4 people proves the original theorem).  
What am I missing here?

Edit:
I think I finally understood the example. My problem was that I was constantly thinking that the two cases are like two sub-theorem that have to be proved in order to prove the 'main' theorem but this is actually not the case.
The real work is done by the 4 sub-cases. These subcases prove the main theorem but in order for them to function they need to be 'supported' by the two cases. 
Special thanks to following people for help:  


*

*@Bram28 - Your comment that I should not focus on the two cases and
focus on the subcases was on the point. I was just blind and could
not understand. 

*@Patrick Stevens - Thanks for your nicely written
example. It is not quite the same as the theorem proof in the
original question but it is a good example nevertheless. 

*@Did - Thank
you for taking the time to review and edit my question so that it is
better.

 A: It's not that it proves the theorem by itself, exactly; it tells you which of two sub-proofs is applicable to any given situation. I'll give an analogy that might be clearer. (The proof is a bit silly because it's so simple, but it should demonstrate the idea.)
We prove that the square of a real is nonnegative. Indeed, there are three cases: $x<0; x=0; x>0$. 
If we specialise to each of those cases, proving the theorem in each limited case, our life is made easier: it is easier than proving the theorem in general, since in each smaller case we know more about $x$.
But you ask, why does this prove the main theorem? Well, any real number must be in one of the three cases above. If someone gives us a real number, we can in principle look at it and see which of the cases it lies in. Then we can perform the steps of the proof which correspond to that case.
For example, here's a toy proof that squares are non-negative: 


*

*If $x=0$ then $x^2=0$ so we're done.

*If $x>0$ then we have the product of two positive numbers (namely $x$ and $x$), which must be positive.

*If $x<0$ then we have the product of two negative numbers, which must be positive.


Why does this prove the theorem? If you give me some $x$ (maybe it's $2$) and ask me to prove that its square is nonnegative, I can look at the cases. I note that $2>0$ so we are in the second case. And then I can apply the proof of the second case to our particular $x$: here, noting that we have the product of two positive numbers (namely 2 and 2), so we're done.
A proof by cases is a collection of several recipes, each one proving the theorem given a bit more information than we started out with. As long as we can demonstrate that we will always have a recipe corresponding to any given example, we can say that the main theorem is proved.

In your theorem, why are the two cases together enough to cover every option? Well, if we're not in the first case, then it's not the case that out of the five others, three have met $x$. So at most two can have met $x$; so at least three must have not. But that tells us we're in the second case! Either we're in the first case or in the second case, so our two recipes will always be enough.
A: Let's call the 2 cases 1 and 2, with sub-cases 1A and 1B, and 2A and 2B respectively.
So then the proof structure looks like this:


*

*Cases: 1 or 2

*$\quad$ Assume case 1

*$\quad $ Cases: 1A or 1B

*$\quad \quad $ Assume Case 1A

*$\quad \quad $ ... There is either a club or a group of 3 => Claim Holds for 1A!

*$\quad \quad $ Assume Case 1B

*$\quad \quad $ ... There is either a club or a group of 3 => Claim Holds for 1B!

*$\quad $ There is either a club or a group of 3 => Claim Holds for 1!

*$\quad $ Assume Case 2

*$\quad$ Cases: 2A or 2B

*$\quad \quad$ Assume 2A

*$\quad \quad$ ... There is either a club or a group of 3 => Claim Holds for 2A!

*$\quad \quad $ Assume 2B

*$\quad \quad $ ... There is either a club or a group of 3 => Claim Holds for 2B!

*$\quad $ There is either a club or a group of 3 => Claim Holds for 2!

*There is either a club or a group of 3 => Claim Holds!
The proof by cases steps are 8, 15, and 16. So at step 8, you say: since we know that A1 or A2 (because we work within assumption A), and since both of those two cases leads to the claim, we know that the claim is true in case A. In step 15, you do the same for case B, on the basis of B1 and B2. And so then at the end (step 16) you say: we know that either A or B. but since both lead to the claim, the claim is always true.
So: you may indeed not immediately see how the claim holds in case A ... Indeed, why do you think they further subdivide it into A1 and A2? It is exactly because it really is not easy to see. but once you have shown it for A1 and A2, you know it is true for A ... Even if that is not immediately clear.
