Do well pruned $\omega_1$ Aronszajn trees have any branches? If a tree is well pruned then any node will have a node above it at any level of the tree above the nodes level. 
Doesn't this mean that no branch can have a countable height, because if you assume a branch has countable height x, then you can find a node above level x that is a part of the branch, so the branch didn't have height x. And since it's a $\omega_1$ Aronszajn tree, it has no uncountable branch.
Something must be wrong with my reasoning, because dependant choice means that the tree must have an infinite branch.
 A: Just because there are no cofinal branches, does not mean there are no branches of countable length. This is just a consequence of Hausdorff's maximality principle: every partial order—in particular trees—have maximal chains, branches in the case of trees. If the tree is an Aronszajn tree, then this branch is countable, and if it is pruned, then there are no leaves, and the branch has an order type of a limit ordinal.
Consider the following tree, as a very simple example: Start with one branch of length $\omega$, then above each node in the branch, split into a branch of length $\omega_1$.
The tree has height $\omega_1$, but the original branch from which we split is still a maximal chain, it is still countable, it is still a pruned tree.
What is crucial in the Aronszajn tree construction is that when you are given a branch, and you walk "upwards" on that branch, then either at some point you have to split away from it, or you'll get stuck at a limit step. This is very apparent when you look at the classic construction with rational sequences: we can walk very far on any given branch, but ultimately we get stuck at a limit step. So in order to walk through all the levels, we always have to split from whatever branch we were "walking on".
