The problem with using match() here is that it is anchored to the beginning of the string. You want to find a single special character anywhere in the string, so use search() instead:

def no_special_characters(s, pat=re.compile('[@_!#$%^&*()<>?/\|}{~:]')):
    if pat.search(s):
        print(s + " has special characters")
    else:
        print(s + " has NO special characters")

You could also keep using match() with the following regex pattern:

.*[@_!#$%^&*()<>?/\|}{~:].*

would you say the below is ok to enter in one special character

Regexp('.*[\¬\!\"\£\$\%\^\&\*\(\)\_\+\`\-\=\{\}\:\@\~\<\>\?\[\]\;\'\#\,\.\/\\\|]'

Yes, your expression will match any string that includes at least one of the specified special characters

[ and ] are treated as meta characters in regex. You'll need to escape them:

In [1]: re.sub(re.compile("<!\[.*?\]>"), "", "<![if support]> hello")
Out[1]: ' hello'

As a simplification (courtesy Wiktor Stribiżew), you can escape just the first left paren, shortening your regex to "<!\[.*?]>".

After the opening square bracket the only special characters are -, ^ and ]:

[|\^&+\-%*/=!>]

You can find the list of special characters here:

• Regular Expression Syntax
• python-regex-cheatsheet

Use re.escape

>>> import re
>>> re.escape(r'\ a.*$')
'\\\\\\ a\\.\\*\\$'
>>> print(re.escape(r'\ a.*$'))
\\\ a\.\*\$
>>> re.escape('www.stackoverflow.com')
'www\\.stackoverflow\\.com'
>>> print(re.escape('www.stackoverflow.com'))
www\.stackoverflow\.com

Repeating it here:

re.escape(string)

Return string with all non-alphanumerics backslashed; this is useful if you want to match an arbitrary literal string that may have regular expression metacharacters in it.

As of Python 3.7 re.escape() was changed to escape only characters which are meaningful to regex operations.