Check whether letters of word are in alphabetical order

CJam, 8 bytes

lel_$_&=

Here is a test harness for all examples in the challenge. This returns 0 or 1 (which are falsy and truthy, respectively, in CJam).

And here is a script to filter the word list in the question (takes a few seconds to run). You'll have to copy the word list into the input field manually, because it's too long for a permalink.

Explanation

l        "Read input.";
 el      "Convert to lower case.";
   _$    "Get a copy and sort it.";
     _&  "Remove duplicates (by computing the set intersection with itself).";
       = "Check for equality with original (lower case) word.";

Regex (any flavor), 55 bytes

Some people don't consider regex to be a programming language, but it's been used before, and it's not close to being the shortest.

^a?b?c?d?e?f?g?h?i?j?k?l?m?n?o?p?q?r?s?t?u?v?w?x?y?z?$

I've added one byte for the i (case-insensitive) flag. This is very straightforward and might be shorter to generate on the fly.

If regex alone are not allowed, you can use this 56-byte Retina program suggested by Martin Büttner:

i`^a?b?c?d?e?f?g?h?i?j?k?l?m?n?o?p?q?r?s?t?u?v?w?x?y?z?$

Running this on the wordlist linked above yielded 10 6-letter words in alphabetical order.

["abhors", "almost", "begins", "begirt", "bijoux", "biopsy", "chimps", "chinos", "chintz", "ghosty"]

Python 3, 44 bytes

*s,=input().lower()
print(sorted(set(s))==s)

A simple approach - check uniqueness, check sortedness.