A thread can not gracefully kill another thread, so with your current code, foo never terminates. (With thread.daemon = True the Python program will exit when only daemon threads are left, but that does not allow you to terminate foo without also terminating the main thread.)
Some people have tried to use signals to halt execution, but this may be unsafe in some cases.
If you can modify foo, there are many solutions possible. For instance, you could check for a threading.Event to break out of the while-loop. 
But if you can not modify foo, you could run it in a subprocess using the multiprocessing module since unlike threads, subprocesses can be terminated. Here is an example of how that might look:
import time
import multiprocessing as mp
def foo(x = 1):
    cnt = 1
    while True:
        time.sleep(1)
        print(x, cnt)
        cnt += 1
def timeout(func, args = (), kwds = {}, timeout = 1, default = None):
    pool = mp.Pool(processes = 1)
    result = pool.apply_async(func, args = args, kwds = kwds)
    try:
        val = result.get(timeout = timeout)
    except mp.TimeoutError:
        pool.terminate()
        return default
    else:
        pool.close()
        pool.join()
        return val
if __name__ == '__main__':
    print(timeout(foo, kwds = {'x': 'Hi'}, timeout = 3, default = 'Bye'))
    print(timeout(foo, args = (2,), timeout = 2, default = 'Sayonara'))
yields
('Hi', 1)
('Hi', 2)
('Hi', 3)
Bye
(2, 1)
(2, 2)
Sayonara
Note that this has some limitations too. 
- subprocesses receive a copy of the parent processes' variables. If you modify a variable in a subprocess, it will NOT affect
the parent process. If your function - funcneeds to modify variables, you will need to use a shared variable.
 
- arguments (passed through - args) and keywords (- kwds) must be
picklable.
 
- processes are more resource-heavy than threads. Usually, you only
want to create a multiprocessing Pool once at the beginning of a
program. This timeoutfunction creates aPoolevery time you call it. This was necessary since we neededpool.terminate()to
terminatefoo. There might be a better way, but I haven't thought of it.