Are there some examples of CAT(1) spaces which are not trees which have disconnected Gromov boundary?

$\begingroup$ Yes: let $T$ be (the 2skeleton of) an equilateral triangle in the hyperbolic plane $H^2$. Consider two copies of $T$ glued on their vertices, and take the universal covering, with the length metric. Then it is obviously QI to a tree and CAT($1$), but not isometric to a tree. $\endgroup$ – YCor Mar 14 '15 at 15:05

$\begingroup$ Second example (if you don't want something quasiisometric to a tree): consider a horodisc in the hyperbolic plane. Then it's CAT($1$) and its Gromov boundary is reduced to a point. $\endgroup$ – YCor Mar 14 '15 at 15:08

$\begingroup$ @YCor How is a point disconnected? $\endgroup$ – Igor Rivin Mar 14 '15 at 16:59

2$\begingroup$ oh, I saw "totally disconnected". Otherwise it's even much easier, just take a wedge of two copies of $H^2$, then the Gromov boundary is a disjoint union of 2 circles. $\endgroup$ – YCor Mar 14 '15 at 23:31

$\begingroup$ This is homework; the question should be moved. As a hint  you should think of examples of CAT(1) spaces and their boundaries. Then think about how you can cut spaces into pieces (or glue spaces together) and how the boundary changes under those operations. $\endgroup$ – Sam Nead Mar 15 '15 at 2:21
Yes, the free product of any two word hyperbolic groups has disconnected Gromov boundary. For proof see the nice survey of KapovichBenakli., section 7.

$\begingroup$ word hyperbolic groups are not exactly CAT(1). (But there are many examples.) $\endgroup$ – Anton Petrunin Mar 14 '15 at 17:19

1$\begingroup$ You need at least one example for which you know that the free product is $CAT(1)$; it's OK for the free product of a surface group with $\mathbf{Z}$. $\endgroup$ – YCor Mar 14 '15 at 23:32
For sure you can not make it to be manifold without boundary.
You can start with a tree and glue to it many compact pieces applying Reshetnyak gluing theorem.