TITLE: Open coloring axiom vs. CH
QUESTION [15 upvotes]: Is there a simple, direct proof that the open coloring axiom contradicts CH (straight from the definitions, no machinery allowed)? The separable metric spaces version of OCA, if that helps.
Edit: probably it wasn't clear what I was asking. Let's say, hypothetically, that I'm writing a book on forcing that is meant to be as elementary and readable as possible. I've just introduced OCA and I have half a page to show the reader that it contradicts CH. Can I do this without quoting external theorems?

REPLY [5 votes]: Another option is to use the same approach as Kunen in his book "Set Theory" (2011). In section V.6 he work with $SOCA$ (a weakening of $OCA_{T}$ that only ask for the existance of one of the homogeneous subsets, not for a covering).
In approximately two pages (364-366) he presents $SOCA$ and shows that the axiom implies $\neg CH$. The way he do it is with weakly Luzin sets.
His proof work as follows: 
1) Define a set in ${R}^{n}$ ($R$ is the reals) as skinny iff the set of directions is not dense in the unit sphere.
2) Defines a weakly Luzin as a set whose all uncountable subsets are not skinny.
3) Define  $A$ as an $\epsilon$-directed set iff there is a point in the unit sphere such that all the directions of $A$ have an angle at most of $\epsilon$ to that point.
4) Notice that every $\epsilon$-directed set is skinny.
5) Using induction and compactness show that $SOCA$ implies that every set is contains an $\epsilon$-directed set.
I was a litle vague because I left the reference. I can try to be more precise with definitions and proofs if needed.