The infinite, binary, rooted tree $T$ is, I believe, countable, because it's isomorphic to the Prufer 2-group $\Bbb Z[\frac12]/\Bbb Z$ which is represented by a subset of the rational numbers.
But what about the infinite, binary directed graph, which is not rooted (but still connected)? So it's the same thing as $T$ everywhere, locally, in that two edges lead into every vertex and one out, but does not converge. Is this countable?
By directed, I mean an antisymmetric order so no vertex is a successor of itself.
Because that's what I'm used to, I will be using the terminology where each node has a single parent node (and a single line of ancestors), one "sister" node, and two children (and a rooted binary tree of successors). Then, assuming that you in your unrooted tree still want any two nodes to have a closest common ancestor a finite distance away, then yes, it is still countable.
Take an arbitrary node $v$ and define a function $f$ on the nodes of your graph by letting $f(u)$ for a vertex $u$ be the distance from $v$ to the closest common ancestor of $u$ and $v$.
$v$ is the root of a subtree of your graph, and for any node in this tree, $f$ has value $0$, as the closest common ancestor of $v$ and any of the successors of $v$ is $v$ itself. So $f^{-1}(0)$ is a binary rooted tree.
$f^{-1}(1)$ contains the direct successor of $v$, as well as any nodes that are the successors of the sister node of $v$, which make up a binary rooted tree.
Similarily, for any integer $n>0$, the set $f^{-1}(n)$ is a binary rooted tree, with an additional single node added to the top. You already know that each of these are countable, and therefore their union is also countable.