You can compare strings directly. x<y means "does x come before y alphabetically?" So you can replace your second block with:

if len(x) == len(y) and x < y:
    return x

This has the advantage of being O(n) (sorting a string is O(n log n)). A character (or string) in Python is "less than" another character if it comes before it in alphabetical order, so in order to see if a string is in alphabetical order we just need to compare each pair of adjacent characters. Also, note that you take range(len(word) - 1) instead of range(len(word)) because otherwise you will overstep the bounds of the string on the last iteration of the loop.

def isInAlphabeticalOrder(word):
    for i in range(len(word) - 1):
        if word[i] > word[i + 1]:
            return False
    return True

I am not 100% sure of this, but your code will probably return unexpected results if you use it with accented characters. accènt will probably return false because è is later in the extended ASCII table than n, but humans will view it as alphabetical because they view è the same as e.

There are libraries that can help you with this. https://github.com/kmike/text-unidecode has a more permissive license. https://pypi.python.org/pypi/Unidecode gives slightly better results and performance, but has a GPL license which is harder to use.

Python's string comparisons are lexical by default, so you should be able to call max and get away with it:

In [15]: sentence
Out[15]: ['this', 'is', 'a', 'sentence']
In [16]: max(sentence)
Out[16]: 'this'

Of course, if you want to do this manually:

In [16]: sentence
Out[16]: ['this', 'is', 'a', 'sentence']

In [17]: answer = ''

In [18]: for word in sentence:
   ....:     if word > answer:
   ....:         answer = word
   ....:         

In [19]: print answer
this

Or you can sort your sentence:

In [20]: sentence
Out[20]: ['this', 'is', 'a', 'sentence']

In [21]: sorted(sentence)[-1]
Out[21]: 'this'

Or, sort it reversed:

In [25]: sentence
Out[25]: ['this', 'is', 'a', 'sentence']

In [26]: sorted(sentence, reverse=True)[0]
Out[26]: 'this'

But if you want to fully manual (which is so painful):

def compare(s1, s2):
    for i,j in zip(s1, s2):
        if ord(i)<ord(j):
            return -1
        elif ord(i)>ord(j):
            return 1
    if len(s1)<len(s2):
        return -1
    elif len(s1)>len(s2):
        return 1
    else return 0

answer = sentence[0]
for word in sentence[1:]:
    if compare(answer, word) == -1:
        answer = word

# answer now contains the biggest word in your sentence

If you want this to be agnostic of capitalization, be sure to call str.lower() on your words first:

sentence = [word.lower() for word in sentence] # do this before running any of the above algorithms

Python can compare strings with < and >: https://docs.python.org/3/tutorial/datastructures.html#comparing-sequences-and-other-types

The problem is you are only printing for the cases of:

• strings are in alphabetical order
• strings are in reverse alphabetical order

Instead, try this:

if name_1 < name_2 and name_2 < name_3:
    print("The names are in alphabetical order.")
else:
    print("The names are not in alphabetical order.")