Choosing a file in Python with simple Dialog

How about using tkinter?

from Tkinter import Tk     # from tkinter import Tk for Python 3.x
from tkinter.filedialog import askopenfilename

Tk().withdraw() # we don't want a full GUI, so keep the root window from appearing
filename = askopenfilename() # show an "Open" dialog box and return the path to the selected file
print(filename)

Done!

Python 3.x version of Etaoin's answer for completeness:

from tkinter.filedialog import askopenfilename
filename = askopenfilename()

With EasyGui:

import easygui
print(easygui.fileopenbox())

To install:

pip install easygui

Demo:

import easygui
easygui.egdemo()

In Python 2 use the tkFileDialog module.

import tkFileDialog

tkFileDialog.askopenfilename()

In Python 3 use the tkinter.filedialog module.

import tkinter.filedialog

tkinter.filedialog.askopenfilename()