It seems that this method was brought up in a bug report here. Some code was written (given here) but unfortunately it doesn't seem that it made it into the final Python 3.4 release.

Incidentally the code that was proposed was extremely similar to the code you have in your question:

# As a method of a Path object
def expanduser(self):
    """ Return a new path with expanded ~ and ~user constructs
    (as returned by os.path.expanduser)
    """
    return self.__class__(os.path.expanduser(str(self)))

EDIT

Here is a rudimentary subclassed version PathTest which subclasses WindowsPath (I'm on a Windows box but you could replace it with PosixPath). It adds a classmethod based on the code that was submitted in the bug report.

from pathlib import WindowsPath
import os.path

class PathTest(WindowsPath):

    def __new__(cls, *args, **kwargs):
        return super(PathTest, cls).__new__(cls, *args, **kwargs)

    @classmethod
    def expanduser(cls):
        """ Return a new path with expanded ~ and ~user constructs
        (as returned by os.path.expanduser)
        """
        return cls(os.path.expanduser('~'))

p = PathTest('C:/')
print(p) # 'C:/'

q = PathTest.expanduser()
print(q) # C:\Users\Username

I'm not quite sure I understand what you're trying to do. However, it's no wonder you're getting stuck in a loop: by re-initializing p.glob() you're starting all over again every time!

p.glob() is actually a generator object, which means it will keep track of its progress by itself. You may just use it the way it was meant to be used: by just iterating over it.

So, for instance, you might be better served by the following:

redpath = os.path.realpath('.')         
thispath = os.path.realpath(redpath)         
thispath = os.path.realpath(thispath+'/../../../..')
p = Path(thispath)
chosen = None
for text_file in p.glob('**/*.fits'):
    print("Is this the correct file path?")
    print(text_file)
    userinput = input("y or n")
    if userinput == 'y':
        chosen = text_file
        break
if chosen:
    print ("You chose: " + str(chosen))

Using os.path

To get the parent directory of the directory containing the script (regardless of the current working directory), you'll need to use __file__.

Inside the script use os.path.abspath(__file__) to obtain the absolute path of the script, and call os.path.dirname twice:

from os.path import dirname, abspath
d = dirname(dirname(abspath(__file__))) # /home/kristina/desire-directory

Basically, you can walk up the directory tree by calling os.path.dirname as many times as needed. Example:

In [4]: from os.path import dirname

In [5]: dirname('/home/kristina/desire-directory/scripts/script.py')
Out[5]: '/home/kristina/desire-directory/scripts'

In [6]: dirname(dirname('/home/kristina/desire-directory/scripts/script.py'))
Out[6]: '/home/kristina/desire-directory'

If you want to get the parent directory of the current working directory, use os.getcwd:

import os
d = os.path.dirname(os.getcwd())

Using pathlib

You could also use the pathlib module (available in Python 3.4 or newer).

Each pathlib.Path instance have the parent attribute referring to the parent directory, as well as the parents attribute, which is a list of ancestors of the path. Path.resolve may be used to obtain the absolute path. It also resolves all symlinks, but you may use Path.absolute instead if that isn't a desired behaviour.

Path(__file__) and Path() represent the script path and the current working directory respectively, therefore in order to get the parent directory of the script directory (regardless of the current working directory) you would use

from pathlib import Path
# `path.parents[1]` is the same as `path.parent.parent`
d = Path(__file__).resolve().parents[1] # Path('/home/kristina/desire-directory')

and to get the parent directory of the current working directory

from pathlib import Path
d = Path().resolve().parent

Note that d is a Path instance, which isn't always handy. You can convert it to str easily when you need it:

In [15]: str(d)
Out[15]: '/home/kristina/desire-directory'