What is wrong with:

if word in mystring: 
   print('success')

Swear = ["curse", "curse", "curse"]

for i in Swear:
    if i in Userinput:
        print 'Quit Cursing!'

You should read up on the differences between lists and tuples.

words = 'blue yellow'

if 'blue' in words:
    print 'yes'
else:
    print 'no'

Also nightly blues would contain blue, but not as a whole word. If this is not what you want, split the wordlist:

if 'blue' in words.split():
    …

This should work - it will ensure only whole words are matched, and sentences must match from the start.

def string_contains(str1,str2):
    lst1 = str1.split(' ')
    lst2 = str2.split(' ')

    if len(lst1) <= len(lst2):
        return lst1 == lst2[:len(lst1)]

    return False

print (string_contains("llo", "hello how are you"))  # False
print (string_contains("hello", "hello how are you"))  # True
print (string_contains("hello how", "hello how are you"))  # True
print (string_contains("hello how a", "hello how are you"))  # False

Basic implementation

Generally, you use split() to split a string of words into a list of them. If the list has more than one element, it's True (i.e. you could print yes)

def contains_multiple_words(s):
  return len(s.split()) > 1

Punctuation support

To support punctuation etc as well, you can split on a regular expression, via the re module's split command:

import re

def contains_multiple_words(s):
  return len(re.compile('\W').split(s)) > 1

The regular expression character class \W means any single non-word character, e.g. punctuation or spaces (see the Python regex syntax guide for details).

Thus, splitting on this instead of just space (the default in the first example) allows for a more realistic idea of "words".