What you're trying to do here is to return a subset of the original mesh (the term segmentation in your original title is a little bit misleading). Trimesh does indeed provide a built-in solution for this:

import trimesh

# Cube mesh 
vertices = [
    [0., 0., 0.], [0., 0., 1.], [0., 1., 0.], [0., 1., 1.],
    [1., 0., 0.], [1., 0., 1.], [1., 1., 0.], [1., 1., 1.]
]
faces = [
    [0, 6, 4], [0, 2, 6], [0, 3, 2], [0, 1, 3],
    [2, 7, 6], [2, 3, 7], [4, 6, 7], [4, 7, 5],
    [0, 4, 5], [0, 5, 1], [1, 5, 7], [1, 7, 3]
]
mesh = trimesh.Trimesh(vertices, faces)
mesh.show()

# Submesh
subfaces = [2, 3]
submesh = mesh.submesh([subfaces], append=True)
submesh.show()

As stated in the Trimesh documentation it requires the indices of the faces you wish to keep.

Remember that your subVertices and subFaces are linked. If you "delete" a vertex you must also delete the faces that contain it (the opposite is not necessarily true). In other words, you have either to use subFaces or deduce the faces that contain your subVertices.

Cheers!