The problem I'm facing is to show there exists a countable anti-chain in an Aronszajn tree $T$. I thought of something, and I wanted to ask you if my proof is correct.
So, the idea is to proceed inductively and in each finite step take a node of the tree which is on some level above the end of every possible branch which goes through any of the previous nodes taken. Such a node has to exist since all the branches are at most countable, so they are bounded in $\omega_1$.
Technically I though we could take $x_0$ to be any node on the $0$-th level and $x_n$ to be some node from a level above $\gamma_{x_{n-1}}>\textrm{length of the longest branch going through any of }x_i\ i<n$. There is no guarantee level $\gamma_{x_{n-1}}$ is non-empty, but some of the levels above surely are, since height of the tree is $\omega_1$.
In each step I make at most countably many steps, so $\{x_n:n\in\omega\}$ is a countable anti-chain.
I feel something's not right. What do you think?